= Solution
If $M$ is <Noetherian module>[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 to
$$
0\longrightarrow P\cap N\longrightarrow P\longrightarrow(P+N)/N\longrightarrow0
$$
shows that $P$ is finitely generated. Thus $M$ is Noetherian.
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