Yes. Take any sequence $\{x_i\}$ for which $F(x_i)$ is bounded. For each $x_i$ consider the sequence satisfying (2), $\{x^\epsilon_i\}$ s.t. $x^\epsilon_i\to x_i$. Due to equicoercivity, $x_i^{1/i}$ has to be in a compact set, so it has a converging subsequence, thus $x_i$ must also have a converging subsequence.

Comment: equicoercivity is essential, e.g., take $F_\epsilon=\epsilon x^2$. $F_\epsilon\stackrel{\Gamma}{\to}0$, which is not coercive.

Yes. Take any sequence $\{x_i\}$ for which $F(x_i)$ is bounded. For each $x_i$ consider the sequence satisfying (2), $\{x^\epsilon_i\}$ s.t. $x^\epsilon_i\to x_i$. Consider $x_i^{\epsilon_i}$ such that $F_{\epsilon_i}(x_i^{\epsilon_i})$ is bounded. Due to equicoercivity, $x_i^{\epsilon_i}$ has to be in a compact set, so it has a converging subsequence, thus $x_i$ must also have a converging subsequence.

Comment: equicoercivity is essential, e.g., take $F_\epsilon=\epsilon x^2$. $F_\epsilon\stackrel{\Gamma}{\to}0$, which is not coercive.