= Solution
A <filtration of a ring> is an exhaustive increasing family of additive subgroups $F_iR$, indexed by the integers, with $1\in F_0R$ and $F_iR\,F_jR\subseteq F_{i+j}R$. A compatible <filtration of a module> on a right <module> satisfies $F_iM\,F_jR\subseteq F_{i+j}M$. The <associated graded module> has components $F_iM/F_{i-1}M$, and its action by the <associated graded ring> is
$$
(m+F_{i-1}M)(r+F_{j-1}R)=mr+F_{i+j-1}M.
$$
Changing either representative changes the product only by an element of $F_{i+j-1}M$, so the action is well-defined.
The <subspace filtration> on $N$ is $F_iN=N\cap F_iM$. The <quotient filtration> is $F_i(M/N)=(F_iM+N)/N$. Inclusion induces an <injective> map $\operatorname{gr}N\to\operatorname{gr}M$, and the quotient induces a <surjective> map to $\operatorname{gr}(M/N)$. If $m\in F_iM$ maps to zero in its degree-$i$ quotient, write $m=n+m'$ with $n\in N$ and $m'\in F_{i-1}M$. Then $n\in N\cap F_iM$, so the kernel is exactly the image of $\operatorname{gr}N$. We have the <short exact sequence>
$$
\boxed{0\longrightarrow\operatorname{gr}N
\longrightarrow\operatorname{gr}M
\longrightarrow\operatorname{gr}(M/N)\longrightarrow0},
$$
which is <exactness of associated graded modules for induced filtrations>.
For the negative case, $F_iR=R$ for $i\geq0$. Completeness means that $R\to\varprojlim_q R/F_{-q}R$ is an <isomorphism>; in particular the <filtration of a ring> is separated. If $x\in F_{-1}R$ and $r\in R$, then $(xr)^q\in F_{-q}R$. The <geometric series>
$$
\sum_{q=0}^{\infty}(xr)^q
$$
converges in the <complete negative filtration> and is a two-sided inverse of $1-xr$, by the finite geometric identity and continuity of multiplication. The <unit criterion for the Jacobson radical> now gives
$$
\boxed{F_{-1}R\subseteq J(R)}.
$$
Let $I$ be any <right ideal>. The <associated graded module> $\operatorname{gr}I$ for its <subspace filtration> is a <homogeneous ideal> on the right in $\operatorname{gr}R$. If $\operatorname{gr}R$ is a <right Noetherian ring>, choose homogeneous generators that lift to $x_1,\ldots,x_s\in I$, with degrees $d_1,\ldots,d_s\leq0$.
For $x\in I\cap F_dR$, cancel its degree-$d$ symbol by a sum $\sum_i x_i a_{i,0}$ with $a_{i,0}\in F_{d-d_i}R$. The remainder belongs to $I\cap F_{d-1}R$. Repeat the cancellation in each succeeding degree. After $q$ steps,
$$
x-\sum_i x_i\sum_{j=0}^{q-1}a_{i,j}\in F_{d-q}R,
\qquad a_{i,j}\in F_{d-j-d_i}R.
$$
For every $i$, the coefficient series converges to some $a_i\in R$ by completeness. Passing to the limit gives $x=\sum_i x_i a_i$. Thus these finitely many $x_i$ generate the actual <right ideal> $I$; no assumption that $I$ is closed is needed. Since every <right ideal> is finitely generated, \b[$R$ is right Noetherian], by <complete negative filtered-graded transfer of Noetherianity>.
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