Why does the vacuum even have permeability and permittivity? The vacuum is empty, yet it seems to have 2 properties: it's absolute permeability and absolute permittivity, which have specific, finite, non-zero values.  Why?
Why are the vacuum permeability and permittivity non-zero and non-infinite?
What would a universe in which the vacuum permeability and permittivity were zero look like?
What would a universe in which the vacuum permeability and permittivity were infinite look like?
 A: Both are meaningless values that go away when you use Gaussian units. It is meaningful that neither is zero and neither is infinite, but their particular finite values don’t have physical significance, just as how the values of other dimensional constants don’t have significance.
A: 
Why are the vacuum permeability and permittivity non-zero and non-infinite?

Why not?

What would a universe in which the vacuum permeability and permittivity were zero look like?

Because $$c = \frac{1}{\sqrt{\epsilon_0 \mu_0}},$$
you'd have an infinite speed of light as Wolphram jonny pointed out.

What would a universe in which the vacuum permeability and permittivity were infinite look like?

Wolphram jonny answered this and it follows from the equation I gave.
A: I will assume throughout this answer that we fix the value of $c$ independently of $\varepsilon_0$ or $\mu_0$. The vacuum permittivity and permeability are related to one another by $\varepsilon_0\mu_0 =  1/c^2$, so they're not independent constants — as we should expect given that electricity and magnetism are both manifestations of the same fundamental force. 
The permittivity is related to the dimensionless fine structure constant $\alpha$ by $\alpha = \frac{1}{4\pi\varepsilon_0} \frac{e^2}{\hbar c}$. The fine structure constant determines the strength of the coupling of charges to the electromagnetic field. Since it's dimensionless, it doesn't depend on a choice of units and in this sense is more fundamental than $\varepsilon_0$.
If we take $\alpha \rightarrow 0$ ($\varepsilon_0 \rightarrow \infty$), charges aren't affected by EM fields at all, and there's no electromagnetic interaction between charges. There would be no atoms, so no macroscopic matter as we know it. If we take $\alpha \rightarrow \infty$ ($\varepsilon_0 \rightarrow 0$), then the EM coupling between charges is infinitely strong. I don't really have good intuition for what happens in this case.
We can see a little physics of both of these limits from Coulomb's law,
\begin{equation}
F = \frac{1}{4\pi\varepsilon_0} \frac{q q'}{r^2},
\end{equation}
where the former limit gives $F\rightarrow 0$ and the latter $F \rightarrow \infty$ for finite charges and distances.
A: If they were zero, you recover Newtonian mechanics (the speed of light would become infinite). If they were infinite, there is no recognizable universe, and you would have frozen light (zero speed) and thus neither mass nor reference frames could exist.
A: For the record, I was able to determine what the consequences for various values of the vacuum permeability and the vacuum permittivity would be by using the following formulas:
$$c_{0} = \sqrt{\frac{1}{\varepsilon_{0}\mu_{0}}}$$
$$\varepsilon_0\mu_0 =  1/c_{0}^2$$
$$Z_{0} = \sqrt{\frac{\mu_{0}}{\varepsilon_{0}}}$$
$$\alpha = \frac{e^{2}Z_{0}}{2h}$$
$$h = \frac{e^{2}Z_{0}}{2\alpha}$$
$$e = \sqrt{\frac{2\alpha h}{Z_{0}}}$$
where

*

*$\varepsilon_{0}$ is the Vacuum Permittivity (a.k.a. the Electric Constant)

*$\mu_{0}$ is the Vacuum Permeability (a.k.a. the Magnetic Constant)

*$c_{0}$ is the Speed of Light in the vacuum

*$Z_{0}$ is the Vacuum Impedance (a.k.a. Impedance of Free Space)

*e is the Elementary Charge

*h is the Planck Constant

*$\alpha$ is the Fine-structure Constant

Depending on which 2 of (e, h, $\alpha$) you consider to be fixed, it has some interesting results.
A: Permeability of free space describes the curvature of the universe, or alternately, the dissipation of an omnidirectional force with distance.
Standard value: μ0 = 4π×10^−7 H/m
The 10^−7 comes from the particular definition of the Ampere. If we "correct" this definition to not contain the 10^−7 embedded in it, we get μ0 = 4π which is merely the surface area of the unit sphere in three (space) dimensions. This corresponds to the inverse square law.
The two-dimensional universe would have a value of μ0 = 2π and the one-dimensional universe would have a value of μ0 = 1.
A four-dimensional universe would have μ0 = 2π^2.
A zero permeability would result in a singularity in the laws of physics that I cannot transform into any meaningful description.
An infinite permeability implies an infinite number of dimensions, which in turn results in forces having no ranges.
In theory, μ0 could be changed with local curvature and be used as the mechanism to adjust the physical laws for highly-curved regions of spacetime. However I know of nobody who actually does this.
