= Solution
{c}
For the <submodule> $N\subseteq M$, define its <subspace filtration> and the <quotient filtration> on the <quotient module> $Q=M/N$ by
$$
N_j=N\cap M_j,\qquad Q_j=(M_j+N)/N.
$$
They are compatible <filtrations of a module>, since multiplication by $F_iR$ sends $N_j$ into $N_{i+j}$ and $Q_j$ into $Q_{i+j}$.
In each degree, inclusion and the quotient map induce
$$
\iota_j:N_j/N_{j-1}\longrightarrow M_j/M_{j-1},\qquad
\pi_j:M_j/M_{j-1}\longrightarrow Q_j/Q_{j-1}.
$$
The first map is <injective>: if $n\in N_j$ belongs to $M_{j-1}$, then $n\in N\cap M_{j-1}=N_{j-1}$. The second map is <surjective> by the definition of $Q_j$.
For exactness in the middle, a class $m+M_{j-1}$ lies in $\ker\pi_j$ precisely when $m\in M_j$ and $m+N\in Q_{j-1}$. This means $m=m'+n$ for $m'\in M_{j-1}$ and $n\in N$. Since $m,m'\in M_j$, one also has $n\in N_j$, and
$$
m+M_{j-1}=n+M_{j-1}\in\operatorname{im}\iota_j.
$$
Conversely, every class from $N_j$ maps to zero in $Q_j/Q_{j-1}$. Taking the <direct sum> over all degrees gives the <short exact sequence> of <graded modules>
$$
0\longrightarrow\operatorname{gr}N\longrightarrow\operatorname{gr}M
\longrightarrow\operatorname{gr}(M/N)\longrightarrow0.
$$
The maps respect the $\operatorname{gr}R$-action because they come from <module homomorphisms> compatible with the <filtrations of a module>. Consequently
$$
\boxed{\operatorname{gr}M/\operatorname{gr}N\cong\operatorname{gr}(M/N)}
$$
as $\operatorname{gr}R$-<modules>. This proves <exactness of associated graded modules for induced filtrations>.
Furthermore $A=\operatorname{gr}R$ is a <Noetherian ring>, and $\operatorname{gr}M$ is a <finitely generated module>. Its <submodule> $\operatorname{gr}N$ and its <quotient module> $\operatorname{gr}(M/N)$ are therefore <finitely generated modules>. Hence the induced <subspace filtration> and <quotient filtration> are both <good filtrations for the order filtration>.
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