= Solution
Set $S=A\setminus\mathfrak m$, a <multiplicative subset> because $\mathfrak m$ is a <prime ideal>. The <quotient module> $B=A/\mathfrak a$ is a <cyclic module> generated by $\bar1$, so its <localization of a module> $S^{-1}B$ is zero precisely when $\bar1/1=0$. By the <vanishing criterion in a module localization>, this is equivalent to the existence of $s\in S$ with $s\bar1=0$ in $B$. The latter condition says $s\in\mathfrak a$. Therefore
$$
\boxed{B_{\mathfrak m}=0\iff\mathfrak a\cap(A\setminus\mathfrak m)\ne\varnothing\iff\mathfrak a\not\subseteq\mathfrak m.}
$$
Explicitly, if $s\in\mathfrak a\setminus\mathfrak m$, multiplication by $s$ annihilates $B$ but becomes invertible after <localization>, forcing every localized element to be zero.
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