I would like to clarify my proof(Sketch) for part A:
We know that $G_{n-1}$ is finite and abelian(i.e. $G_{n-1}/\{e\}$ is abelian ) therefore it is solvable. Since, $G_{n-2}/G_{n-1}$ is finite and abelian it is also solvable. Therefore, $G_{n-2}$ is solvable (Since $G_{n-1}$ is normal in $G_{n-2}$). Continuing in this manner we can show that $G_1$ and $G_0/G_1$ is solvable therefore G is solvable. Is this valid?
This is perfectly valid. But it may not be enough, depending on what the exercise assume known. That is, you are using the fact that for $N \subseteq G$ a normal subgroup, $N$ solvable and $G/N$ solvable imply that $G$ is. This is true, but if you are not sure why, you should write a proof.