Here's a claim from which all these statements would follow: For any Cauchy sequence $r_i \in R$, and $x \in M$, the sequence $r_i x$ is eventually constant. (The reason for believing this claim is that, for example, for the first part of i) we have to take a "limit" of $r_i x$ while remaining in $M$, and it's hard to see how else one could do this with no topology on $M$.)
Can we prove this claim? First, let's try this: Since $m$ is an associated prime of $M$, we know that for some $x \in M$, the annihilator of $x$ is $m$. For such an $x$, we can see that $r_i x$ is eventually constant because the Cauchyness of the $r_i$ implies that for some large $N$, we have $r_i - r_j \in m$ for all $i,j \ge N$, hence $r_i x = r_j x$ for all $i,j \ge N$. Unfortunately, this argument doesn't cover all $x \in M$.
However, for a general $x$, if we can show that ann($x$) contains a power of $m$, then we could still apply the same argument as above. Somehow this should follow from the fact that $m$ is the only associated prime ideal of $M$ (and we may have to assume that $R$ is Noetherian). Edit: See YACP's answer for a proof that ann($x$) contains a power of $m$.