Volume per $k$-state We're talking about Fermi-energies for the first time, for $N$ spin $\frac{1}{2}$ particles in a 3D box, and she writes down that $$2 \times \frac{1}{8} \times \frac{4}{3} \pi k_f^3 = Nq \times \frac{\pi^2}{V}.$$  She says that the $2$ comes from the fact that there are $2$ spin states per $k$ (which I get), the $\frac{1}{8}$ because we only want to take the first octant of the sphere (fine) and the $\frac{4}{3} \pi k_f^3$ comes from the volume of the sphere filled states, which I understand since $$E\propto k_x^2+k_y^2+k_z^2$$ for an infinite well (a sphere for constant $E$). She says that the $\frac{\pi^2}{V}$ is the volume per $k$-state? I don't understand where she got that from.  And now she's asking us to repeat the process for a 3D oscillator, but for a 3D oscillator, isn't $E \propto (n_x +n_y + n_z)$? That won't have a volume; it's just a plane, right? Help!
 A: I will explain your issue with the infinite well problem first, and then  give you a hint for the harmonic oscillator, since this looks like a homework question.
Infinite Potential Well:
Remember that in an infinite (3D) cubic potential well, the momentum eigenvalues are restricted to be of the following form because of the boundary conditions:
$$\mathbf k_\mathbf n = \frac{\pi}{L}\mathbf n$$
where $\mathbf n$ is a vector with natural numbers as its components, i.e. $\mathbf n\in \mathbb N^3$. Now the energy levels are also given by:
$$E_{\mathbf n} = \frac{\hbar^2}{2m}|\mathbf k_\mathbf n|^2$$
So now you want to place your spin $1/2$ fermions on each energy level according to the Pauli principle, and see where the last ($N^\mathrm{th}$) fermion is going to end up. 
Imagine the three dimensional $\mathbf k$-space, and keep in mind that not every point in this space is a valid momentum; the momentum always has to be of the form $\mathbf k = \frac{\pi}{L}\mathbf n$. This means that the locus of valid momenta in $\mathbf k$-space is a grid with adjacent gridlines being $\pi/L$ 
 far apart. And since each vertex corresponds to a distinct state, the $\mathbf k$-space volume corresponding to each state (i.e. the volume of each small cube in the grid) is $(\pi/L)^3=\pi^3/V$. Also note that since each component of $\mathbf n$ is positive, the relevant part of $\mathbf k$-space for us is simply the first octant.
Now the magnitude of the momentum $k_f$ corresponding to the last fermion describes (one eighth of) a sphere as mentioned in the question. So now if we calculate the total volume inside this sphere, which is the volume corresponding to states with $|\mathbf k|\leq k_f$, we get:
$$(\text{volume in $\mathbf k$-space with $|\mathbf k|\leq k_f$}) = \frac18 \times \frac43 \pi k_f^3$$
On the other hand, the total number of fermions should be twice the number of states with $|\mathbf k|\leq k_f$ (twice because of the Pauli principle). In other words:
$$N=2\times(\text{# of states with $|\mathbf k|\leq k_f$})$$
$$N=2\frac{\text{volume in $\mathbf k$-space with $|\mathbf k|\leq k_f$}}{\text{volume in $\mathbf k$-space per state}}=2\frac{\frac18 \times \frac43 \pi k_f^3}{\pi^3/V}$$
$$N=\frac{Vk_f^3}{3\pi^2}$$
Finally, this gives:
$$k_f = \Big(\frac{3\pi^2N}{V}\Big)^{1/3}$$
Notice that a somewhat less confusing way of doing this would be to work directly in terms of $\mathbf n$ instead of $\mathbf k$. The eigenenergies are simply $E_{\mathbf n} = \frac{\hbar^2}{2m}|\frac{\pi}{L}\mathbf n|^2$. You can simply do the same argument in "$n$-space" instead; which is more straightforward since the distance between adjacent gridlines is $1$, so that the n-space volume per state is also simply $1$. Analogous to the previous method, this gives:
$$N=2\frac{\text{volume in $\mathbf n$-space with $|\mathbf n|\leq n_f$}}{\text{volume in $\mathbf n$-space per state}}=2\frac{\frac18 \times \frac43 \pi n_f^3}{1}=\frac13 \pi n_f^3$$
Solving for $n_f$ and using $k_f=\pi n_f/L$ gives the same answer as before.
3D Harmonic Oscillator
For the harmonic oscillator, the energy levels are given by:
$$E_{n_xn_yn_z}=\hbar \omega(\frac{3}{2}+n_x+n_y+n_z)$$
with $n_x,n_y,n_z\in \{0,1,2,...\}$. Again, imagine your "$n$-space". Valid states in this space are again described by a grid for which the vertices are simply points with integer non-negative $n$ values. The equation $E_f = \hbar \omega(\frac{3}{2}+n_x+n_y+n_z)$ describes a plane as you've already figured out. Simply count the the number of states, i.e. the volume in $n$-space (of what shape?), for which $E_{n_xn_yn_z} \leq E_f$ (drawing a diagram helps!), and equate it to twice your total number of fermions to find your final answer.
