If $\mathcal{C}$ is a category, is there a right adjoint for the sections functor $\Gamma(U,\_):\mathbf{Psh}_{\mathcal{C}}\to\mathbf{Sets},$ where $\mathbf{Psh}_{\mathcal{C}}$ is the category of presheaves of sets over $\mathcal{C}$?
For presheaves with values in abelian groups I built it sending an abelian group $A$ to the "skyscraper" presheaf $A_U$ such that $A_U(U)=A$ and $A_U(W)=0$ for every other object $W$.
If I try to transfer this argument on presheaves with values in $\mathbf{Sets}$ however I would have to replace $0$ with the singleton $\{*\}$, and it seems to not work anymore, because there are several maps from $\{*\}$ to $A$ and I have not a canonical choice for a "skyscraper" presheaf.
Edit: As explained in the comments below, my definition for presheaves with values in abelian groups is incomplete and it seems that there is not a good way to produce an adjoint from it.
When we replace $\mathcal{C}$ by its dual, your question is just if for every object $A \in \mathcal{C}$ the evaluation functor $\mathrm{ev}_A : \mathrm{Hom}(\mathcal{C},\mathbf{Set}) \to \mathbf{Set}$ has a right adjoint. This is indeed the case, we define a functor $R : \mathbf{Set} \to \mathrm{Hom}(\mathcal{C},\mathbf{Set})$ by $R(X)(B) = X^{\mathrm{Hom}(B,A)}$. Then for functors $F : \mathcal{C} \to \mathbf{Set}$ and sets $X$ we have $$\begin{align*} \mathrm{Hom}(F,R(X)) & \cong \int_{B \in \mathcal{C}} \mathrm{Hom}(F(B),X^{\mathrm{Hom}(B,A)}) \\ & \cong \int_{B \in \mathcal{C}} \mathrm{Hom}(\mathrm{Hom}(B,A),\mathrm{Hom}(F(B),X)) \\ & \cong \mathrm{Hom}(F(A),X). \quad (\text{Yoneda}) \end{align*}$$ This adjunction can also be seen in the framework of Kan extensions, since $\mathrm{ev}_A$ identifies with the pullback functor $A^*$ for the evident functor from the terminal category $A : 1 \to \mathcal{C}$. Explicitly, the right Kan extension maps a set $X$ to the functor $\mathrm{Ran}_{A}(X) : \mathcal{C} \to \mathbf{Set}$ defined by $\mathrm{Ran}_A(X)(B) := \lim_{\varphi : B \to A} X = X^{\mathrm{Hom}(B,A)}.$
You can replace $ \mathbf{Set}$ by any category with powers.