Can we compute all digits of the Euler-Mascheroni constant?

In practice , we would be screwed in the case the decimal expansion is terminating. Nevertheless, $\gamma$ is computable also in this case since if it has a terminating decimal expansion, it is rational and then there is an algorithm to determine all digits. That would not help us since we would not know whether this is actually the case.

The difficulty you describe is known as the equality problem. It is in general not possible to decide the equality of two expressions if they can be "complicated enough". Not sure whether the well known limit for $\gamma$ falls into this category.

That the decimal expansion of $\gamma$ terminates is even less likely than that $\gamma$ is rational, but as far as I know, we cannot rule it out.

Yes, if you look at the series expansion. https://en.wikipedia.org/wiki/Euler%27s_constant#Series_expansions

One can approximate the constant by something which is easy to compute $- \log(n) + \sum_{k=1}^n \frac{1}{k}$. There is also a guarantee about how fast this converges to $\gamma$ as $n$ tends to infinity. Using that guarantee you can be sure about the first digits. It is proven that the denominator would have to huge if $\gamma$ is rational.