Proof.
Parts (1), (2), (3) are local on $S$ hence we may and do assume $S$ is affine. Since $a$ is quasi-compact we conclude that $X$ is quasi-compact. Since $\mathcal{F}^\bullet $ is locally bounded, we conclude that $\mathcal{F}^\bullet $ is bounded.
For (1) and (2) we can use the first spectral sequence $R^ pa_*\mathcal{F}^ q \Rightarrow R^{p + q}a_*\mathcal{F}^\bullet $ of Derived Categories, Lemma 13.21.3. Combining Cohomology of Schemes, Lemma 30.4.5 and Homology, Lemma 12.24.11 we conclude.
Let us prove (3) by the induction principle of Cohomology of Schemes, Lemma 30.4.1. Namely, for a quasi-compact open of $U$ of $X$ consider the condition that $R(a|_ U)_*(\mathcal{F}^\bullet |_ U)$ has finite tor dimension. If $U, V$ are quasi-compact open in $X$, then we have a relative Mayer-Vietoris distinguished triangle
\[ R(a|_{U \cup V})_*\mathcal{F}^\bullet |_{U \cup V} \to R(a|_ U)_*\mathcal{F}^\bullet |_ U \oplus R(a|_ V)_*\mathcal{F}^\bullet |_ V \to R(a|_{U \cap V})_*\mathcal{F}^\bullet |_{U \cap V} \to \]
by Cohomology, Lemma 20.33.5. By the behaviour of tor amplitude in distinguished triangles (see Cohomology, Lemma 20.48.6) we see that if we know the result for $U$, $V$, $U \cap V$, then the result holds for $U \cup V$. This reduces us to the case where $X$ is affine. In this case we have
\[ Ra_*\mathcal{F}^\bullet = a_*\mathcal{F}^\bullet \]
by Leray's acyclicity lemma (Derived Categories, Lemma 13.16.7) and the vanishing of higher direct images of quasi-coherent modules under an affine morphism (Cohomology of Schemes, Lemma 30.2.3). Since $\mathcal{F}^ n$ is $S$-flat by assumption and $X$ affine, the modules $a_*\mathcal{F}^ n$ are flat for all $n$. Hence $a_*\mathcal{F}^\bullet $ is a bounded complex of flat $\mathcal{O}_ S$-modules and hence has finite tor dimension.
Proof of part (5). Denote $a' : (X, a^{-1}\mathcal{O}_ S) \to (S, \mathcal{O}_ S)$ the obvious flat morphism of ringed spaces. Part (5) says that
\[ K \otimes _{\mathcal{O}_ S}^\mathbf {L} Ra'_*\mathcal{F}^\bullet = Ra'_*(L(a')^*K \otimes _{a^{-1}\mathcal{O}_ S}^\mathbf {L} \mathcal{F}^\bullet ) \]
Thus Cohomology, Equation (20.54.2.1) gives a functorial map from the left to the right and we want to show this map is an isomorphism. This question is local on $S$ hence we may and do assume $S$ is affine. The rest of the proof is exactly the same as the proof of Lemma 36.22.1 except that we have to show that the functor $K \mapsto Ra'_*(L(a')^*K \otimes _{a^{-1}\mathcal{O}_ S}^\mathbf {L} \mathcal{F}^\bullet )$ commutes with direct sums. This is where we will use $\mathcal{F}^ n$ has the structure of a quasi-coherent $\mathcal{O}_ X$-module. Namely, observe that $K \mapsto L(a')^*K \otimes _{a^{-1}\mathcal{O}_ S}^\mathbf {L} \mathcal{F}^\bullet $ commutes with arbitrary direct sums. Next, if $\mathcal{F}^\bullet $ consists of a single quasi-coherent $\mathcal{O}_ X$-module $\mathcal{F}^\bullet = \mathcal{F}^ n[-n]$ then we have $L(a')^*G \otimes _{a^{-1}\mathcal{O}_ S}^\mathbf {L} \mathcal{F}^\bullet = La^*K \otimes _{\mathcal{O}_ X}^\mathbf {L} \mathcal{F}^ n[-n]$, see Cohomology, Lemma 20.27.4. Hence in this case the commutation with direct sums follows from Lemma 36.4.5. Now, in general, since $S$ is affine (hence $X$ quasi-compact) and $\mathcal{F}^\bullet $ is locally bounded, we see that
\[ \mathcal{F}^\bullet = (\mathcal{F}^ a \to \ldots \to \mathcal{F}^ b) \]
is bounded. Arguing by induction on $b - a$ and considering the distinguished triangle
\[ \mathcal{F}^ b[-b] \to (\mathcal{F}^ a \to \ldots \to \mathcal{F}^ b) \to (\mathcal{F}^ a \to \ldots \to \mathcal{F}^{b - 1}) \to \mathcal{F}^ b[-b + 1] \]
the proof of this part is finished. Some details omitted.
Proof of part (4). Let $a' : (X, a^{-1}\mathcal{O}_ S) \to (S, \mathcal{O}_ S)$ be as above. Since $\mathcal{F}^\bullet $ is a locally bounded complex of flat $a^{-1}\mathcal{O}_ S$-modules we see the complex $a^{-1}\mathcal{G} \otimes _{a^{-1}\mathcal{O}_ S} \mathcal{F}^\bullet $ represents $L(a')^*\mathcal{G} \otimes _{a^{-1}\mathcal{O}_ S}^\mathbf {L} \mathcal{F}^\bullet $ in $D(a^{-1}\mathcal{O}_ S)$. Hence (4) follows from (5).
$\square$
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