 Solution

If $M$ is Noetherian, every submodule of $N$ is a submodule of $M$, and submodules of $M/N$ correspond to submodules of $M$ containing $N$; hence both are Noetherian.

Conversely, suppose $N$ and $M/N$ are Noetherian. For any $P\subseteq M$, the intersection $P\cap N$ is finitely generated and the image $(P+N)/N$ is finitely generated. Lifting generators of the image and applying part i toshows that $P$ is finitely generated. Thus $M$ is Noetherian.