Assume (2). Consider the object $R\SheafHom(M, \mathcal{O})$ and the composition map $$ R\SheafHom(M, \mathcal{O}) \otimes_\mathcal{O}^\mathbf{L} M \to \mathcal{O} $$ To prove this is an isomorphism, we may work locally. Thus we may assume $\mathcal{O} = \prod_{a \leq n \leq b} \mathcal{O}_n$ and $M = \bigoplus_{a \leq n \leq b} \mathcal{H}^n[-n]$. Then it suffices to show that $$ R\SheafHom(\mathcal{H}^m, \mathcal{O}) \otimes_\mathcal{O}^\mathbf{L} \mathcal{H}^n $$ is zero if $n \not = m$ and equal to $\mathcal{O}_n$ if $n = m$. The case $n \not = m$ follows from the fact that $\mathcal{O}_n$ and $\mathcal{O}_m$ are flat $\mathcal{O}$-algebras with $\mathcal{O}_n \otimes_\mathcal{O} \mathcal{O}_m = 0$. Using the local structure of invertible $\mathcal{O}$-modules (Modules on Sites, Lemma \ref{sites-modules-lemma-invertible}) and working locally the isomorphism in case $n = m$ follows in a straightforward manner; we omit the details. Because $D(\mathcal{O})$ is symmetric monoidal, we conclude that $M$ is invertible. \medskip\noindent Assume (1). The description in (2) shows that we have a candidate for $\mathcal{O}_n$, namely, $\SheafHom_\mathcal{O}(H^n(M), H^n(M))$. If this is a locally finite family of sheaves of rings and if $\mathcal{O} = \prod \mathcal{O}_n$, then we immediately obtain the direct sum decomposition $M = \bigoplus H^n(M)[-n]$ using the idempotents in $\mathcal{O}$ coming from the product decomposition. This shows that in order to prove (2) we may work locally in the following sense. Let $U$ be an object of $\mathcal{C}$. We have to show there exists a covering $\{U_i \to U\}$ of $U$ such that with $\mathcal{O}_n$ as above we have the statements above and those of (2) after restriction to $\mathcal{C}/U_i$. Thus we may assume $\mathcal{C}$ has a final object $X$ and during the proof of (2) we may finitely many times replace $X$ by the members of a covering of $X$. \medskip\noindent Choose an object $N$ of $D(\mathcal{O})$ and an isomorphism $M \otimes_\mathcal{O}^\mathbf{L} N \cong \mathcal{O}$. Then $N$ is a left dual for $M$ in the monoidal category $D(\mathcal{O})$ and we conclude that $M$ is perfect by Lemma \ref{lemma-left-dual-derived}. By symmetry we see that $N$ is perfect. After replacing $X$ by the members of a covering, we may assume $M$ and $N$ are represented by a strictly perfect complexes $\mathcal{E}^\bullet$ and $\mathcal{F}^\bullet$. Then $M \otimes_\mathcal{O}^\mathbf{L} N$ is represented by $\text{Tot}(\mathcal{E}^\bullet \otimes_\mathcal{O} \mathcal{F}^\bullet)$. After replacing $X$ by the members of a covering of $X$ we may assume the mutually inverse isomorphisms $\mathcal{O} \to M \otimes_\mathcal{O}^\mathbf{L} N$ and $M \otimes_\mathcal{O}^\mathbf{L} N \to \mathcal{O}$ are given by maps of complexes $$ \alpha : \mathcal{O} \to \text{Tot}(\mathcal{E}^\bullet \otimes_\mathcal{O} \mathcal{F}^\bullet) \quad\text{and}\quad \beta : \text{Tot}(\mathcal{E}^\bullet \otimes_\mathcal{O} \mathcal{F}^\bullet) \to \mathcal{O} $$ See Lemma \ref{lemma-local-actual}. Then $\beta \circ \alpha = 1$ as maps of complexes and $\alpha \circ \beta = 1$ as a morphism in $D(\mathcal{O})$. After replacing $X$ by the members of a covering of $X$ we may assume the composition $\alpha \circ \beta$ is homotopic to $1$ by some homotopy $\theta$ with components $$ \theta^n : \text{Tot}^n(\mathcal{E}^\bullet \otimes_\mathcal{O} \mathcal{F}^\bullet) \to \text{Tot}^{n - 1}( \mathcal{E}^\bullet \otimes_\mathcal{O} \mathcal{F}^\bullet) $$ by the same lemma as before. Set $R = \Gamma(\mathcal{C}, \mathcal{O})$. By Lemma \ref{lemma-category-summands-finite-free} we find that we obtain \begin{enumerate} \item $M^\bullet = \Gamma(X, \mathcal{E}^\bullet)$ is a bounded complex of finite projective $R$-modules, \item $N^\bullet = \Gamma(X, \mathcal{F}^\bullet)$ is a bounded complex of finite projective $R$-modules, \item $\alpha$ and $\beta$ correspond to maps of complexes $a : R \to \text{Tot}(M^\bullet \otimes_R N^\bullet)$ and $b : \text{Tot}(M^\bullet \otimes_R N^\bullet) \to R$, \item $\theta^n$ corresponds to a map $h^n : \text{Tot}^n(M^\bullet \otimes_R N^\bullet) \to \text{Tot}^{n - 1}(M^\bullet \otimes_R N^\bullet)$, and \item $b \circ a = 1$ and $b \circ a - 1 = dh + hd$, \end{enumerate} It follows that $M^\bullet$ and $N^\bullet$ define mutually inverse objects of $D(R)$. By More on Algebra, Lemma \ref{more-algebra-lemma-invertible-derived} we find a product decomposition $R = \prod_{a \leq n \leq b} R_n$ and invertible $R_n$-modules $H^n$ such that $M^\bullet \cong \bigoplus_{a \leq n \leq b} H^n[-n]$. This isomorphism in $D(R)$ can be lifted to an morphism $$ \bigoplus H^n[-n] \longrightarrow M^\bullet $$ of complexes because each $H^n$ is projective as an $R$-module. Correspondingly, using Lemma \ref{lemma-category-summands-finite-free} again, we obtain an morphism $$ \bigoplus H^n \otimes_R \mathcal{O}[-n] \to \mathcal{E}^\bullet $$ which is an isomorphism in $D(\mathcal{O})$. Here $M \otimes_R \mathcal{O}$ denotes the functor from finite projective $R$-modules to $\mathcal{O}$-modules constructed in the proof of Lemma \ref{lemma-category-summands-finite-free}. Setting $\mathcal{O}_n = R_n \otimes_R \mathcal{O}$ we conclude (2) is true. \medskip\noindent If $(\mathcal{C}, \mathcal{O})$ is a locally ringed site, then given an object $U$ and a finite product decomposition $\mathcal{O}|_U = \prod_{a \leq n \leq b} \mathcal{O}_n|_U$ we can find a covering $\{U_i \to U\}$ such that for every $i$ there is at most one $n$ with $\mathcal{O}_n|_{U_i}$ nonzero. This follows readily from part (2) of Modules on Sites, Lemma \ref{sites-modules-lemma-locally-ringed} and the definition of locally ringed sites as given in Modules on Sites, Definition \ref{sites-modules-definition-locally-ringed}. From this the implication (2) $\Rightarrow$ (3) is easily seen. The implication (3) $\Rightarrow$ (2) holds without any assumptions on the ringed site. We omit the details.