\DeclareMathOperator{\Ass}{Ass}

Yes. Here’s a proof.

First, note that since $\sqrt{I} \neq m$, we have $I : m^\infty \neq R$, so that $R/(I : m^\infty) \neq 0$.

Now suppose every element of $m$ is a zero-divisor on $R/(I : m^\infty)$. Then $m \subseteq \bigcup \Ass (R/(I : m^\infty))$, so by prime avoidance $m \in \Ass (R/(I: m^\infty))$. Thus there is some $x\in R \setminus (I : m^\infty)$ with $m x \subseteq I : m^\infty$. That is, there is some $t$ with $mx \subseteq I : m^t$. Thus, $x \in ((I : m^t) : m) = (I : m^{t+1}) \subseteq (I : m^\infty)$, which is a contradiction.

Yes. Here’s a proof.

First, note that since $\sqrt{I} \neq m$, we have $I : m^\infty \neq R$, so that $R/(I : m^\infty) \neq 0$.

Now suppose every element of $m$ is a zero-divisor on $R/(I : m^\infty)$. Then $m \subseteq \bigcup \operatorname{Ass} (R/(I : m^\infty))$, so by prime avoidance $m \in \operatorname{Ass} (R/(I: m^\infty))$. Thus there is some $x\in R \setminus (I : m^\infty)$ with $m x \subseteq I : m^\infty$. That is, there is some $t$ with $mx \subseteq I : m^t$. Thus, $x \in ((I : m^t) : m) = (I : m^{t+1}) \subseteq (I : m^\infty)$, which is a contradiction.