**Proof.**
The first statement follows from the second, hence we only prove (2). We will prove this by induction on the length of the complex $\mathcal{E}^\bullet $. If $\mathcal{E}^\bullet \cong \mathcal{E}[-n]$ for some direct summand $\mathcal{E}$ of a finite free $\mathcal{O}$-module and integer $n \geq a$, then the result follows from Lemma 21.44.5 and the fact that $\mathcal{F}^{n - 1} \to \mathop{\mathrm{Ker}}(\mathcal{F}^ n \to \mathcal{F}^{n + 1})$ is surjective by the assumed vanishing of $H^ n(\mathcal{F}^\bullet )$. If $\mathcal{E}^ i$ is zero except for $i \in [a, b]$, then we have a split exact sequence of complexes

\[ 0 \to \mathcal{E}^ b[-b] \to \mathcal{E}^\bullet \to \sigma _{\leq b - 1}\mathcal{E}^\bullet \to 0 \]

which determines a distinguished triangle in $K(\mathcal{O}_ U)$. Hence an exact sequence

\[ \mathop{\mathrm{Hom}}\nolimits _{K(\mathcal{O}_ U)}( \sigma _{\leq b - 1}\mathcal{E}^\bullet , \mathcal{F}^\bullet ) \to \mathop{\mathrm{Hom}}\nolimits _{K(\mathcal{O}_ U)}(\mathcal{E}^\bullet , \mathcal{F}^\bullet ) \to \mathop{\mathrm{Hom}}\nolimits _{K(\mathcal{O}_ U)}(\mathcal{E}^ b[-b], \mathcal{F}^\bullet ) \]

by the axioms of triangulated categories. The composition $\mathcal{E}^ b[-b] \to \mathcal{F}^\bullet $ is homotopic to zero on the members of a covering of $U$ by the above, whence we may assume our map comes from an element in the left hand side of the displayed exact sequence above. This element is zero on the members of a covering of $U$ by induction hypothesis.
$\square$

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