Let $G$ be a PRG. Are the following $G'$ constructions based on $G$ necessarily PRGs?
Looks good to me! My intuition is that the PRG property means "random in, random out".

In 1. the input is not random so we get no guarantees - indeed, formally one could take any PRG H with 2s-bit inputs  and then define G to be H except on strings where the last s bits are 0, in which case G outputs the all-zero string. Now G is still a PRG, because the probability of hitting one of those modified points is at most $2^{-s}$ which makes a negligible difference overall. But $G'$ is now the function that always outputs the all-zero string.

What I've added here is the step from the intuition to a formal proof: I've given a case where $G$ is provably a PRG (ok, I didn't prove it here, but that's an easy exercise) and $G'$ is provably not a PRG.

In 2. if $s_1 \ldots s_n$ is uniform in $\{0,1\}^n$ then $s_1 \ldots s_{n/2}$ is certainly uniform in $\{0,1\}^{n/2}$. So applying $G$ to this string produces something uniformly distributed in the range of $G$, hence the whole thing is a PRG, just like you said.