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.
To expand on my comments, you need to define how $G$ is extended for inputs of different sizes before you can determine whether the resulting $G'$ are PRG or not.

1. With the definition in your answer that $G(s\_1 || s\_2) = G(s\_1)||G(s\_2)$ it is indeed not a PRG, but you *could* define e.g. $G(s\_1 || s\_2) = G(s\_1 \oplus s\_2)$, in which case $G'$ *is* PRG.
2. Again, it depends on the definition. Suppose $G(t\_1 || t\_2) = H(t\_1)||H\_(t\_2)$ for some PRG $H$ with a smaller input size. Now $G$ is PRG for random inputs $s = t\_1 || t\_2$. Define $G(t\_1) = H(t\_1)$ and $G'$ is PRG. However, if you define $G(t\_1) = G(t\_1 || 0^{|t\_1|})$ you are back to earlier point and $G'$ is *not* PRG.