**Proof.**
Part (1) means $\mathit{QCoh}(\mathcal{O}_ X)$ (a) has all colimits, (b) filtered colimits are exact, and (c) has a generator, see Injectives, Section 19.10. By Schemes, Section 26.24 colimits in $\mathit{QCoh}(\mathcal{O}_ X)$ exist and agree with colimits in $\textit{Mod}(\mathcal{O}_ X)$. By Modules, Lemma 17.3.2 filtered colimits are exact. Hence (a) and (b) hold. To construct a generator $U$, pick a cardinal $\kappa $ as in Lemma 28.23.3. Pick a collection $(\mathcal{F}_ t)_{t \in T}$ of $\kappa $-generated quasi-coherent sheaves as in Lemma 28.23.2. Set $U = \bigoplus _{t \in T} \mathcal{F}_ t$. Since every object of $\mathit{QCoh}(\mathcal{O}_ X)$ is a filtered colimit of $\kappa $-generated quasi-coherent modules, i.e., of objects isomorphic to $\mathcal{F}_ t$, it is clear that $U$ is a generator. The assertions on limits and injectives hold in any Grothendieck abelian category, see Injectives, Theorem 19.11.7 and Lemma 19.13.2.

Proof of (2). To construct $Q$ we use the following general procedure. Given an object $\mathcal{F}$ of $\textit{Mod}(\mathcal{O}_ X)$ we consider the functor

\[ \mathit{QCoh}(\mathcal{O}_ X)^{opp} \longrightarrow \textit{Sets},\quad \mathcal{G} \longmapsto \mathop{\mathrm{Hom}}\nolimits _ X(\mathcal{G}, \mathcal{F}) \]

This functor transforms colimits into limits, hence is representable, see Injectives, Lemma 19.13.1. Thus there exists a quasi-coherent sheaf $Q(\mathcal{F})$ and a functorial isomorphism $\mathop{\mathrm{Hom}}\nolimits _ X(\mathcal{G}, \mathcal{F}) = \mathop{\mathrm{Hom}}\nolimits _ X(\mathcal{G}, Q(\mathcal{F}))$ for $\mathcal{G}$ in $\mathit{QCoh}(\mathcal{O}_ X)$. By the Yoneda lemma (Categories, Lemma 4.3.5) the construction $\mathcal{F} \leadsto Q(\mathcal{F})$ is functorial in $\mathcal{F}$. By construction $Q$ is a right adjoint to the inclusion functor. The fact that $Q(\mathcal{F}) \to \mathcal{F}$ is an isomorphism when $\mathcal{F}$ is quasi-coherent is a formal consequence of the fact that the inclusion functor $\mathit{QCoh}(\mathcal{O}_ X) \to \textit{Mod}(\mathcal{O}_ X)$ is fully faithful.
$\square$

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