= Solution
{c}
Fix a <coherent sheaf> $F$ and an integer $k>M$. Set $K_0=F$. Repeatedly apply <3/a> to construct <short exact sequences of sheaves>
$$
0\longrightarrow K_{j+1}\longrightarrow G_j\longrightarrow K_j\longrightarrow0,
$$
where each $G_j$ is a <coherent sheaf> and a <locally free sheaf>. Each <kernel sheaf> $K_{j+1}$ is a <coherent sheaf> because $X$ is a <noetherian scheme>. Thus the construction can be repeated as many times as needed; it does not require the <kernel sheaves> themselves to be <locally free sheaves>.
For every integer $q>M$, the <long exact sequence in sheaf cohomology> contains
$$
H^q(X,G_j)\longrightarrow H^q(X,K_j)\xrightarrow{\ \delta_j\ }H^{q+1}(X,K_{j+1})\longrightarrow H^{q+1}(X,G_j).
$$
Both outer groups vanish by the hypothesis, since $q$ and $q+1$ exceed $M$. Hence $\delta_j$ is an <isomorphism>. Iterating this <dimension shifting in sheaf cohomology> gives
$$
H^k(X,F)\cong H^{k+1}(X,K_1)\cong\cdots\cong H^{k+s}(X,K_s).
$$
Choose $s\geq0$ with $k+s>N$, where $N$ is the uniform bound from <3/b>. The <coherent sheaf> $K_s$ is a <quasi-coherent sheaf>, so <3/b> makes the last group zero. Therefore
$$
\boxed{H^k(X,F)=0\quad\text{for every coherent sheaf }F\text{ and every }k>M.}
$$
The argument uses a finite initial segment of a <locally free resolution>. There is no need to assume that every <coherent sheaf> has a finite <locally free resolution>: the uniform bound from <3/b> is what terminates the argument.
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