Can a single piece of paper really block all non-electromagnetic ionizing radiation? So in my twelfth grade (about 2 years ago), I had a bit of nuclear physics and in it, we talked about how alpha particles could be blocked by a piece of paper, beta emissions could be blocked by a sheet of aluminium and gamma radiation could only be stopped with sheets of lead and concrete and they gave a very specific diameter like 6mm of aluminium would stop beta radiation or a sheet of A4 paper would stop alpha emissions.
Although I couldn't my books, I found a picture which stereotypically depicts what I'm talking about, on the left, top to bottom, it says "Radiation Alpha, Radiation Beta, Radiation Gamma" and on the bottom, left to right, it says "Paper, Metalic sheet, Concrete wall"

My question is: in high exposure does this simple theory hold true? (I am  90% convinced it does not). As in, if you get to crazy activity or exposure to radiation, a shred of paper won't protect you from alpha particles or that even if you put a certain amount of concrete or lead with enough gamma emissions eventually it won't isolate entirely from ionizing radiation
I would like to know how can helium molecules go past a solid material or how gamma rays eventually "penetrate through lead", as well as the formulas behind it.
Side question for this one respond in the comments if you feel like it, since I don't deem this one to be a whole question to put on Physics SE, to get a more solid footing into nuclear physics specifically fission, fusion and atom interactions what steps should I take? Especially the formulas and theory.
 A: Alpha particles, that is, Helium nuclei, are very much different to gamma rays, in that alpha particles have mass and charge. Gamma rays are electromagnetic rays and are hence massless and carry no charge. This means that gamma rays will not interact with other materials as much, and can quite easily pass through many materials without interacting or colliding with other atoms. This means that their ability to penetrate materials is vastly greater than that for alpha rays, as well as beta rays.
Alpha particles are massive and they are "larger" than gamma rays and  are also much more massive (and larger) than beta rays$^1$, and they have twice the magnitude of charge than for beta rays, so they are more likely to collide as well as Coulomb interact with other atoms that make up a material. Again keeping in mind that gamma rays have neither charge or mass, so they will interact much less with matter than for alpha and beta rays.
As for an equation, the Bethe-Bloch equation "describes the mean energy loss per distance travelled of swift charged particles (protons, alpha particles, atomic ions) traversing matter".
Equations for gamma rays (they have energy that depends on their frequency), since they interact with matter via different mechanisms, cross-sections for each of the mechanisms must be computed and this is  detailed here.

in high Exposure does this simple theory hold true?

Cosmic rays which are mostly protons and alpha particles, are ultra "high-energy protons and atomic nuclei that move through space at nearly the speed of light." In fact, these rays can have a penetration depth in the order 100-1,000 more than average beta rays and gamma rays.
So of course, if you have a very highly energetic beam of alpha particles, compared to a much weaker beam of beta rays  and gamma rays, it is likely that the alpha rays will be more penetrating. This means that your assumption is correct.
Also, in the equations cited in the links above, you can see that the energy loss per unit depth is dependent on the incident (original energy before striking material) energy of the beam. In such cases, it may very well be that such highly energetic beams will require more than a sheet of paper.
The same could said about highly energetic gamma rays requiring thicker lead or concrete to increase the level of blocking. But the diagrams and information you have provided are generally true, in that they describe "usual energies" for such forms of radiation. But you are in fact correct in that the information you provided does not necessarily hold true for unusually high energy alpha rays, and gamma rays.
The exposure, and I take it that by this you mean the amount of incident particles, will not affect their penetration depth, as this is a property of individual particles.
But certainly if you mean human exposure, then a greater influx of radiation can be more harmful than a smaller flux, simply because a greater region is being exposed. The more cells that are struck, the more biological damage.
$^1$ Alpha particles are Helium nuclei, $He^{++}$ and have mass about four atomic mass units, where $$1\ amu \approx 1.66 \times 10^{-27} kg$$ where an amu is about the mass of a proton (or neutron).
Its charge is twice that for an electron. Note that beta rays can be positrons or electrons. An electron itself has charge $$q_e \approx -1.6 \times 10^{-19} C$$ which means the charge of an alpha particle $$q_{\alpha} \approx +3.2 \times 10^{-19}C$$
Alpha particles are about $2000$ more massive than beta particles.
A: This source essentially answers your questions.
Alpha and beta particles are all blocked while gamma particles are not. This is because alpha particles and beta particles are electrically charged, and therefore is affected by the electric fields in the matter. Gamma particles are not charged, and attenuates through.

The interactions of the various radiations with matter are unique and determine their penetrability through matter and, consequently, the type and amount of shielding needed for radiation protection. Being electrically neutral, the interaction of gamma rays with matter is a statistical process and depends on the nature of the absorber as well as the energy of the gamma. There is always a finite probability for a gamma to penetrate a given thickness of absorbing material and so, unlike the charged particulate radiations which have a maximum range in the absorber where all are stopped regardless of source strength, some gammas will always get through and, given a strong enough source, a lot may get through.

Emphasis mine.
A decent analogy is shooting a bunch of bullets into a wall. No matter how many bullets you shoot, as long as the walls are decently thick, the bullets won't get through. However, just like shooting bullets in a wall, higher energy bullets can go farther.

For low density materials, the range of 5.5 MeV alphas (from Am-241) is between 4.5 to 5 mg/cm$^2$; higher density materials give a range between 5 and 12 mg/cm$^2$.

Note that the value is given in terms of density times distance of penetration. Accordingly, it penetrates roughly 50 $\mu$m in a piece of paper.
Beta particles penetrates farther.

A useful rule-of-thumb for the maximum range of electrons is that the range (in [mg/cm$^2$]) is half the maximum energy (in MeV).

A 2.3 MeV beta particle beam can penetrate roughly 4.2 mm of aluminum.

Gamma ray interactions with matter are entirely different from that of charged particles. The lack of charge eliminates Coulomb interactions and allows gamma rays to be much more penetrating. The interactions that do occur are by way of the photoelectric effect, Compton scattering, and pair production. The probability for any of these happening is specified by a cross section, and the linear attenuation coefficients for gamma rays are defined by these cross sections.
Since linear attenuation coefficients vary with the density of the absorber, even for the same absorber material, the mass attenuation coefficient μ/ρ (linear attenuation coefficient μ in 1/cm$^-1$ divided by the density ρ in g/cm$^3$) is more useful, and the attenuation law is written as
$I = I_o e^{-(μ/ρ)ρt}$  Equation (1)
where I is the intensity of the radiation and t is the thickness. The product ρt is the significant parameter and the units (as with β and α particles) are mg/cm2, making the exponent in Equation (1) dimensionless.

For gamma particles, we can talk in terms of half value thickness. In other words, we talk about how thick does a material has to be to block half the radiation. Doubling the thickness will ensure only 1/4 of the radiation gets through, and so on.
For Co-60, you'll need roughly 1 cm of lead to block half of the radiation.
A: For alpha particles, the ones we are usually thinking about are ones produced by radioactive decay.
When an helium-4 nucleus with much more energy is talked about, it is often described as a "cosmic ray" instead.
The energy of such particles falls in a narrow band; most in a factor of 2x compared to each other.
Wikipedia graph of alpha particle energy.
The ability of an alpha particle to penetrate depends on its individual energy.  Adding more alpha particles (up to a ridiculous limit) does not make them much better at penetrating to a great extent.
Alpha particles don't keep going until they hit something generally; this isn't a "50% of particles are absorbed every micron" situation.  If that was the case, then every doubling of Alpha particle count would increase penetration by a fixed amount.
Instead, they engage in multiple "small" interactions, each time losing a fraction of their energy.  So after a certain depth, they have lost all of their energy, and they come to a stop.
Bragg curve of energy loss of alpha particles in air over 4 cm or so.
So doubling the amount of alpha particles won't boost penetration depth to any significant (to your health) amount.
Boosting the energy of each alpha particle would increase penetration depth, but even that isn't going to be linear; a higher energy particle may shed more energy per collision.
Because what we usually call alpha particles are helium-4 nucleus within a relatively narrow energy band per particle, the penetration depth doesn't change based on how many we are exposed to.  When the helium-4 nucleus leaves this energy band, we start calling it by a different name (we don't call helium-4 gas a "mixture of low energy alpha particles and beta particles" other than as a joke; cosmic rays we might sometimes refer to as alpha particles, but that still isn't what "piece of paper" blocking refers to.)

As to how things penetrate, solids are at the energy levels of ionizing radiation mostly empty space.  Charged ionizing radiation's interacts with the charged parts of solids and energy exchange occurs.  When the charged ionizing radiation particle's individual energy levels fall sufficiently, the solid becomes less "transparent" and more "occupied".
Uncharged ionizing radiation is less likely to interact, so is more likely to fly right through the material.  The "cross section" of collision is smaller.
The danger of alpha particle emitters is that if they are dust in the air, you get them in your lungs, and then the alpha radiation gets inside your body.  And then it kills you, because it doesn't matter if it only causes damage a few cells deep.
A: So a complete answer to your question requires the PDB (https://pdg.lbl.gov/2019/reviews/rpp2018-rev-passage-particles-matter.pdf), anything less will leave stuff out. It is very difficult to make general statements that don't have exceptions. Nevertheless, there are some salient features that can help understand the passage of particles through matter, i.e.: radiation.
One: Is the radiation charged? Charges particles interact according to the aforementioned Bethe-Bloch equation. That equation describes energy loss due to interactions with atomic electrons. A charged particle losses energy bit-by-bit by ionizing atoms in the shielding.
All that matters is the number of atomic electrons between you and the source. With that, aluminum is the perfect beta shield. It is 1/4 the density of lead, but has 1/3 the electron density, and thus provides sufficient shielding electrons per shielding mass. The 1st figure shows the stopping power per areal density vs. $Z$.

The next question is: How fast is the particle? The two key features of the Bethe Bloch eq. are 1) the minimum ionization (circa $\gamma=4$) and the logarithmic rise there after, and 2) the $1/v^2$ behavior at low energy. The energy losses vs $\gamma$ is shown in the second figure.

That means slow particles are stopped on the spot, and alpha particles are slow. It's 4 nucleons ($4m_p\approx 4\,$GeV/c), while the kinetic energy is only a few MeV.
The final feature of the BB formula is that stopping power goes as the square of the incident particle's charge. Alpha's have 4x the interaction of a proton.
The key take away is that charged particle interaction is characterized by $dE/dX$ (energy loss per distance/density traversed), and each material has a know "stopping power".
Neutral particles are different. They don't see Avogadro's number of atomic electrons that they simply can't avoid losing a little energy to each one they encounter. The continue along at full energy, until they really hit something.
Gammas are neutral photons, they don't see charge, until they do, at which point they interact via the photoelectric effect, Compton scattering, bremsstrahlung, or pair production, dumping a huge fraction of their energy into secondary particles in a single event. This process is probabilistic, and is characterized by exponential attenuation in shielding. (Hence, there's always a probability energy can get through).
What photons need to interact is strong electric field, and the place to find that is near a high Z nucleus: a small volume full of charge. Lead offers the cheapest way to pack high Z into a small volume. In critical applications, tungsten and depleted uranium can be used.
The shielding is characterized by the "radiation length" ($X_0$), which is the range at which a $\gamma$ will lose $1/e$ of its energy, on average. Typically:
$$\frac 1 {X_0}= 716.4\,{\rm g\,cm}^{-2}\frac{A}{Z(Z+1)\ln{\frac{287}{\sqrt Z}}}$$
so you really benefit from high-Z nuclei, of which lead is the cheapest. Detector shielding may involve tens of tons of material...making gold and platinum a no-go.
Last is the neutron, which is not only neutral, it doesn't interact electromagnetically at nuclear energies. The best way to think about it is that it is billiard ball knocking other billiard balls around. If the target is a $A=208$ lead nucleus, it losses almost no energy. Lead is terrible neutron shield. If the target is a proton in $H_2O$, then it shares its energy with a proton and continues on. With that, it can create many ionizing protons as it bounces around your body. Water is good  shielding, wax is also good: the protons in hydrocarbons slow neutrons down. If the wax is borated, the boron captures the neutrons, and that's that.
