I'll write the two lemmas I have questions about, and then ask my questions. For reference, I'm using the following definition of Gorenstein:

$\mathbf{Definition\ 1.15}$ A local noetherian ring $A$ is called Gorenstein if $A$ as a module over itself has a finite injective resolution.

Moreover, I am assuming, as stated, that this implies that $A$ is a dualizing complex for itself, so that the natural map

$M \rightarrow \mathbf{R}Hom^.(\mathbf{R}Hom^.(M,A),A)$

is an isomorphism.

$\mathbf{Lemma\ 1.19}$ Let $X$ be Gorenstein (and noetherian, finite krull dimension, separated, and coh(X) has enough injectives). The the following conditions on a coherent sheaf $\mathscr{F}$ are equivalent.

1. The sheaves $\underline{Ext}^i(\mathscr{F}, \mathscr{O}_x)$ are trivial for all i>0.

2. There is a right locally free resolution

$0 \rightarrow \mathscr{F} \rightarrow Q^0 \rightarrow Q^1 \dots$

$\mathbf{Proof} $ $(ii \implies i)$ Consider a brutal truncation $\sigma^{\leq k} Q^.$ for sufficiently large $k$. Denote by $\mathscr{G}$ the nontrivial cohomology $H^k(\sigma^{\leq k} Q^.)$. For any $i>0$ we have isomorphisms

$\underline{Ext}^i(\mathscr{F}, \mathscr{O}_X) \cong \underline{Ext}^{i+k+1}(\mathscr{G}, \mathscr{O}_X)=0$,

where the last equality comes from the fact that we choose $k$ large, and the remark that since $X$ is Gorenstein, there exists an integer $n_0$ such that $\underline{Ext}^i(\mathscr{F}, \mathscr{O}_X)=0$ for all quasi-coherent $\mathscr{F}$ and all $i>n_0$.

$\mathbf{Question}$: Why can we make any comparison between $\mathscr{F}$ and $\mathscr{G}$? I understand that in the category $D_{Sg}(X)$ we may be able to get an isomorphism $\mathscr{F} \simeq \mathscr{G}[-k-1]$, which would resolve this problem, but here we are just working in the category of chains of complexes of coherent sheaves, no?

Second, related, question-

$\mathbf{Proposition\ 1.23}$ Let $X$ be as above and Gorenstein.Then any object $A \in D_{Sg}(X)$ is isomorphic to the image of a coherent sheaf $\mathscr{F}$ such that $\underline{Ext}^i(\mathscr{F}, \mathscr{O}_X)=0$ for all $i>0$.

Note: My question does not actually use any information about D_{Sg}(X), and can be understood without knowledge of it

$\mathbf{Proof}$: An object $A$ is a bounded complex of coherent sheaves. Let us take a locally free bounded above resolution $P^. \xrightarrow A$ which we know exists. Consider a brutal truncation $\sigma^{\geq -k}P^.$ for sufficiently large $k>0$. Denote by $\mathscr{G}$ the cohomology $H^{-k}(\sigma^{\geq -k}P^.)$. Since $A$ is bounded and $X$ is Gorenstein we know that the comples $\mathbf{R}Hom^.(A, \mathscr{O}_X)$ is bounded. This implies that if $k>>0$ then $\underline{Ext}^i(\mathscr{G}, \mathscr{O}_X)=0$ for all $i>0$. (...)

$\mathbf{Question}$: I am content with absolutely everything in this proof except for the last line. If we compute $\underline{Ext}^i(\mathscr{G}, \mathscr{O}_X)$ as $H^i(Hom^.(\mathscr{G}, I^.)$ (where I let $I^.$ be a right, locally free, finite resolution of $\mathscr{O}_X$, which exists by $X$ being Gorenstein), then I cannot see where $k$ even plays a role. Is there some property of truncating these complexes far enough out (where the complexes ought to have 0 cohomology groups) such that there can be no nonzero maps into the locally free $I^j$ sheaves?

I'll write the two lemmas I have questions about, and then ask my questions. For reference, I'm using the following definition of Gorenstein:

$\mathbf{Definition\ 1.15}$ A local noetherian ring $A$ is called Gorenstein if $A$ as a module over itself has a finite injective resolution.

Moreover, I am assuming, as stated, that this implies that $A$ is a dualizing complex for itself, so that the natural map

$M \rightarrow \mathbf{R}Hom^.(\mathbf{R}Hom^.(M,A),A)$

is an isomorphism.

$\mathbf{Lemma\ 1.19}$ Let $X$ be Gorenstein (and noetherian, finite krull dimension, separated, and coh(X) has enough injectives). The the following conditions on a coherent sheaf $\mathscr{F}$ are equivalent.

1. The sheaves $\underline{Ext}^i(\mathscr{F}, \mathscr{O}_x)$ are trivial for all i>0.

2. There is a right locally free resolution

$0 \rightarrow \mathscr{F} \rightarrow Q^0 \rightarrow Q^1 \dots$

$\mathbf{Proof} $ $(ii \implies i)$ Consider a brutal truncation $\sigma^{\leq k} Q^.$ for sufficiently large $k$. Denote by $\mathscr{G}$ the nontrivial cohomology $H^k(\sigma^{\leq k} Q^.)$. For any $i>0$ we have isomorphisms

$\underline{Ext}^i(\mathscr{F}, \mathscr{O}_X) \cong \underline{Ext}^{i+k+1}(\mathscr{G}, \mathscr{O}_X)=0$,

where the last equality comes from the fact that we choose $k$ large, and the remark that since $X$ is Gorenstein, there exists an integer $n_0$ such that $\underline{Ext}^i(\mathscr{F}, \mathscr{O}_X)=0$ for all quasi-coherent $\mathscr{F}$ and all $i>n_0$.

$\mathbf{Question}$: Why can we make any comparison between $\mathscr{F}$ and $\mathscr{G}$? I understand that in the category $D_{Sg}(X)$ we may be able to get an isomorphism $\mathscr{F} \simeq \mathscr{G}[-k-1]$, which would resolve this problem, but here we are just working in the category of chains of complexes of coherent sheaves, no?

Second, related, question-

$\mathbf{Proposition\ 1.23}$ Let $X$ be as above and Gorenstein.Then any object $A \in D_{Sg}(X)$ is isomorphic to the image of a coherent sheaf $\mathscr{F}$ such that $\underline{Ext}^i(\mathscr{F}, \mathscr{O}_X)=0$ for all $i>0$.

Note: My question does not actually use any information about $D_{Sg}(X)$, and can be understood without knowledge of it

$\mathbf{Proof}$: An object $A$ is a bounded complex of coherent sheaves. Let us take a locally free bounded above resolution $P^. \rightarrow A^.$ which we know exists. Consider a brutal truncation $\sigma^{\geq -k}P^.$ for sufficiently large $k>0$. Denote by $\mathscr{G}$ the cohomology $H^{-k}(\sigma^{\geq -k}P^.)$. Since $A$ is bounded and $X$ is Gorenstein we know that the comples $\mathbf{R}Hom^.(A, \mathscr{O}_X)$ is bounded. This implies that if $k>>0$ then $\underline{Ext}^i(\mathscr{G}, \mathscr{O}_X)=0$ for all $i>0$. (...)

$\mathbf{Question}$: I am content with absolutely everything in this proof except for the last line. If we compute $\underline{Ext}^i(\mathscr{G}, \mathscr{O}_X)$ as $H^i(Hom^.(\mathscr{G}, I^.)$ (where I let $I^.$ be a right, locally free, finite resolution of $\mathscr{O}_X$, which exists by $X$ being Gorenstein), then I cannot see where $k$ even plays a role. Is there some property of truncating these complexes far enough out (where the complexes ought to have 0 cohomology groups) such that there can be no nonzero maps into the locally free $I^j$ sheaves?