Let $M$ be a finitely generated module over a Noetherian ring $R$. I need to show that for a multiplicately closed subset $U\subset R$, $$\bigcap_{P\in \operatorname{Ass}(M)\\ P\cap U=\emptyset}\ker(M\to M_P)=\ker(M\to U^{-1}M)$$ and that either expression equals $\{m\in M:I^nm=0 \text{ for } n>>0\}$ for some ideal $I\subset R$.
1) Suppose $x\in\ker(M\to U^{-1}M) $. Then $ux=0$ in $R$ for some $u\in U$. For any prime ideal $P$ not meeting $U$, the image $x/1$ of $x$ in $M_P$ is zero because $ux=0$ and $u\in R-P$. Thus $x\in \bigcap_{P\in \operatorname{Ass}(M)\\ P\cap U=\emptyset}\ker(M\to M_P)$.
2) Next, Let $x\in \bigcap_{P\in \operatorname{Ass}(M)\\ P\cap U=\emptyset}\ker(M\to M_P)$. Then For every $P\in \operatorname{Ass}M)$ with $P\cap U=\emptyset$ there exists $s_P\in R-P$ such that $s_Px=0$ in $R$. To show that $x\in\ker(M\to U^{-1}M) $ I need to find $u\in U$ such that $ux=0$. How to cook up such a $u$ (from the $s_P$, I guess)?
3) Here I'm also not sure what ideal this should be.
$\newcommand\Ass{\operatorname{Ass}}\newcommand\inv{^{-1}}\newcommand\ann{\operatorname{ann}}$Recall that $\Ass M[U\inv]=\{P\in\Ass M : P\cap U =\varnothing\}$. See Stacks.
Part 2
To show that $$\bigcap_{P\in \Ass M[U\inv]} \ker M\to M_P\subseteq \ker M\to M[U\inv],$$ let $m\in \ker M\to M_P$ for all $P\in\Ass M[U\inv]$. Thus $\ann m \not \subseteq P$ for any $P\in \Ass M[U\inv]$, so by prime avoidance, we can choose $r\in R$ with $rm=0$, but $r\not\in P$ for any $P$ associated to $M[U\inv]$. Thus $r$ is a non-zerodivisor on $M[U\inv]$, which implies that since $rm=0$ in $M$ (and thus in $M[U\inv]$) that $m=0$ in $M[U\inv]$, as desired.
Part 3
I'm slightly confused about what part 3 is asking. Is it saying that there exists some ideal $I\subseteq R$ such that $m\in\ker M\to M[U\inv]$ if and only if there exists $n$ such that $I^nm=0$? If so, I don't have any good ideas right now, but it's a bit late here.