Some questions on Hartshorne I.7: intersections in projective space I am reading I.7 of Hartshorne, and here are some questions I don't understand.

1) Prop. 7.4. Let $M$ be a finitely generated graded module over a noetherian graded ring $S$. Then there exists a filtration $0=M^0\subseteq M^1\subseteq\cdots \subseteq M^r=M$ by graded submodules, such that for each $i$, $M^i/M^{i-1}\cong (S/p_i)(l_i)$, where $p_i$ is a homogeneous prime ideal of $S$, and $l_i\in \mathbb{Z}$.

First, I don't understand why "the zero module admit such a filtration".
Second, why "it is clear that $p\supseteq Ann(M)\Longleftrightarrow p\supseteq Ann(M^i/M^{i-1})$ for some $i$" ?  

2) Proof of Thm 7.5. If $0\rightarrow M'\rightarrow M\rightarrow M''\rightarrow 0$ is a short exact sequence of finitely generated $S=k[x_0,\dots,x_n]$-modules, then $$Z(Ann M)=Z(Ann M')\cup Z(Ann M'').$$

I can't see why.
 A: 1) I believe the statement should be "for each $i > 0$, $M^i/M^{i-1} \cong (S/p_i)(l_i)$" in which case the zero module has a length zero filtration and satisfies this vacuously.
2) If $0 \to M' \to M \to M'' \to 0$ is a short exact sequence of f.g. modules over any commutative ring, then $0 \to (M')_p \to M_p \to (M'')_p \to 0$ is exact for any prime ideal $p$, so $M_p = 0$ iff $(M')_p = 0$ and $(M'')_p = 0$, or equivalently, $M_p \ne 0$ iff $(M')_p \ne 0$ or $(M'')_p \ne 0$. Now use the fact that for f.g. $M$, $Z(\text{ann}(M)) = \{p \in \text{Spec}(R) \mid M_p \ne 0\}$.
3) Applying the reasoning in (2) repeatedly to the short exact sequences 
$$0 \to M^{i-1} \to M^i \to M^i/M^{i-1} \to 0$$ 
shows that $Z(\text{ann}(M)) = \bigcup_{i=1}^n Z(\text{ann}(M^i/M^{i-1}))$.
A: Here is another proof that if 
$$0 \to M' \to M \to M'' \to 0$$ is
s.e.s. of $R$-modules, then a prime ideal $p$ contains
$Ann(M)$ iff it contains one of $Ann(M')$ or $Ann(M'')$.
To see this, let $I' = Ann(M'),$ $I = Ann(M),$ and $I'' = Ann(M'')$.
Then one sees directly, just from the definition of annihilator and
s.e.s., that   $I \subset I' \cap I''$, and that
$I' I'' \subset I.$
Thus, if $p$ is prime and contains $I$, then we see $p$ contains $I' I''$,
and so contains one of $I'$ or $I''$.  Conversely, if $p$ contains 
$I'$ or $I'',$ then it contains $I\cap I''$, and so contains $I$.
