**Proof.**
Part (2) follows trivially from part (1). In the proof of (1) we will use the equivalence $D(A) = D_\mathit{QCoh}(X)$ of Lemma 36.3.5 without further mention. Assume $M^\bullet $ has tor amplitude in $[a, b]$. Then $K^\bullet $ is isomorphic in $D(A)$ to a complex $K^\bullet $ of flat $A$-modules with $K^ i = 0$ for $i \not\in [a, b]$, see More on Algebra, Lemma 15.65.3. Then $E$ is isomorphic to $\widetilde{K^\bullet }$. Since each $\widetilde{K^ i}$ is a flat $\mathcal{O}_ X$-module, we see that $E$ has tor amplitude in $[a, b]$ by Cohomology, Lemma 20.45.3.

Assume that $E$ has tor amplitude in $[a, b]$. Then $E$ is bounded whence $M^\bullet $ is in $K^-(A)$. Thus we may replace $M^\bullet $ by a bounded above complex of $A$-modules. We may even choose a projective resolution and assume that $M^\bullet $ is a bounded above complex of free $A$-modules. Then for any $A$-module $N$ we have

\[ E \otimes _{\mathcal{O}_ X}^\mathbf {L} \widetilde{N} \cong \widetilde{M^\bullet } \otimes _{\mathcal{O}_ X}^\mathbf {L} \widetilde{N} \cong \widetilde{M^\bullet \otimes _ A N} \]

in $D(\mathcal{O}_ X)$. Thus the vanishing of cohomology sheaves of the left hand side implies $M^\bullet $ has tor amplitude in $[a, b]$.
$\square$

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