Is there an energy density limit in GR?

The answer is NO. There is no energy density limit (for all three questions).

The easiest way to see this is that the energy density is just the $T^{00}$ component of the stress energy tensor. The solution in GR depends on the full stress energy tensor, so it is not enough to just talk about the energy density. Furthermore, because the energy density is just a component of a tensor, it is a coordinate system dependent quantity. So starting from a solution that doesn't become a blackhole, and has some energy somewhere, we can always choose the coordinate system to make the energy density arbitrarily large.

More clearly stated: Local Lorentz symmetry alone is enough to show that the energy density is not limited in GR. And furthermore since there exist non-zero energy solutions that don't become blackholes, this also answers your second question.

To make the answer to the third question more clear, let's discuss an exact solution. Consider the Robertson-Walker solution with a perfect fluid. Here's an example stress energy tensor for a perfect fluid in the comoving frame:

$T^{ab} =\left( \begin{matrix} \rho & 0 & 0 & 0 \\ 0 & p & 0 & 0 \\ 0 & 0 & p & 0 \\ 0 & 0 & 0 & p \end{matrix} \right)$

Now if we change to a different coordinate system, using the coordinate transformation: $\Lambda^{\mu}{}_{\nu} =\left( \begin{matrix} \gamma &-\beta \gamma & 0 & 0 \\ -\beta \gamma&\gamma&0&0\\ 0&0&1&0\\ 0&0&0&1\\ \end{matrix} \right)$

We see the energy density will transform as: $\rho' = \gamma^2 \rho + p \beta^2 \gamma^2 = \gamma^2 (\rho + p \beta^2)$

So not only can the energy density be arbitrarily large, but even over a finite volume.