Let $T$ be a completely separable topological space, is $2^T$ still completely separable? I seriously doubt that the subset of $2^T$ cannot be written as the countable union of the basis open sets of $2^T$.
Second question: Let $X$ be a connected separable topological space, is $2^X$ still connected separable? How to prove this?
Not a homework.
If the Vietoris topology on the hyperspace is meant: No for completely separable (a.k.a second countable, $C_2$ or $w(X) = \aleph_0$) as the hyperspace of a countable discrete space is regular, but not normal (and so cannot be second countable as a regular second countable space is normal).
$X$ connected and $T_1$ implies $2^X$ connected (classical, not too hard), and $X$ separable and $T_1$ implies that $2^X$ is separable (the finite subsets of the dense subsets are dense). So yes for $T_1$ spaces (which is often assumed in this hyperspace context to ensure that there are enough closed sets and so that $X$ embeds into $2^X$ naturally via $x \to \{x\}$).