Question: $G$ is a PRG where $|G(s)| > 2 \cdot |s|$.

1. $G'(s) = G(s0^{|s|})$.
2. $G'(s) = G(s\_1...s\_{n/2})$, where $s = s\_1...s\_n$.

My question here is whether my own proposed solutions are correct and do make sense.

Answer: 
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.
In the first case, I would say that $G'$ is not necessarily a PRG because, following from the definition of a PRG, the input seed $w = s0^{|s|}$ to $G$ is not random, so the output can't possibly be pseudorandom: the leftmost $n = |s|$ bits of $w$ correspond to a random seed $s$, but the rightmost $n$ bits are all fixed to $0$. Furthermore, the $G(x)||G(y)$ concatenation $- \enspace x,y \leftarrow \{{0,1}\}^n$, $x \neq y \enspace -$ corresponds to the concatenation of 2 random looking strings; however, the $G(x')||G(y')$ concatenation, where $x' = s$ and $y' = 0^n$, doesn't correspond to the concatenation of 2 random looking strings, as the input seed $y'$ is not random.

In the second case, $G'$ could be a PRG because, following from the definition of a PRG: (1) [**expansion**] $|G(s\_1...s\_{n/2})| > 2 \cdot n/2 = n$, so $|G'(s)| > |s|$; (2) [**pseudo-randomness**] if $s$ is a randomly chosen seed, and $G$ is a PRG, then $s\_1...s\_{n/2}$ is a random string and $G(s\_1...s\_{n/2})$ is indistinguishable from a truly random string $y \leftarrow \{{0,1}\}^{l(n)}$, where $l(n) = |G'(s)|$.

In case there are any mistakes or not clear enough, please specify where am I wrong and argue why.