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Over a commutative unital ring, an ideal maximal among annihilators of nonzero elements of a module is a prime ideal. Each such ideal is the annihilator of a module given by the cyclic submodule generated by that element. If $P=\mathrm{Ann}(m)$ and $ab\in P$ with $b\notin P$, then $bm\ne 0$ and $P\subseteq \mathrm{Ann}(bm)$. Maximality makes these annihilators equal, so $a\in P$. A Noetherian ring ensures that a nonzero module has a maximal element annihilator by the ascending chain condition on ideals.

No. Take the Noetherian ring $R=k[x]$, the finitely generated $R$-moduleand $N=0$. Its annihilator of a module iswhich is a prime ideal and hence a primary ideal. But with $r=x$ and $m=(0,\overline{1})$ we have $m\ne 0$ and $rm=0$, while $x^{k}M\ne 0$ for every $k$ because the free summand survives. Thus $N$ is not a primary submodule.