Is the tensor product of two projective sheaves projective?

The tensor product of two abelian sheaves $\mathcal{F}$ and $\mathcal{G}$ is defined to be the sheafification of $U \mapsto \mathcal{F}(U) \otimes \mathcal{G}(U)$. In the category of abelian groups, the tensor product of projectives is projective, so it seems logical to hope a similar thing with sheaves.

I was trying to look at the presheaf $U \mapsto \mathcal{F} (U) \otimes \mathcal{G}(U)$. It is projective at every section and I was trying to use this to assemble a natural transformation to complete the necessary diagram that would make $\mathcal{F} \otimes \mathcal{G}$ projective, but I don't see a way of doing this that will make the necessary diagrams commutative to assemble a natural transformation.

Another attempt was using the adjunction with the internal sheaf hom because we have that

$$Hom(\mathcal{F}, \mathscr{H}\kern -.5pt om (\mathcal{G}, -)) \cong Hom (\mathcal{F} \otimes \mathcal{G}, -)$$

and $\mathcal{F} \otimes \mathcal{G}$ is projective if and only if $Hom(\mathcal{F} \otimes \mathcal{G}, -)$ is exact. The usual argument with groups is composing exact functors, since the internal hom coincides with the usual hom. However, we don't know here if $\mathscr{H}\kern -.5pt om(\mathcal{G},-)$ is exact when $\mathcal{G}$ is projective (and my attempts at proving this have been unsuccessful as well).

I would also be interested to know if this statement is true in a general closed monoidal category, but the specific application I was looking for is in the category of sheaves.

EDIT: this question is more specifically referring to condensed abelian groups (sheafs of abelian groups on the site of profinite sets), but I thought it might be true in a more general sheaf setting.

This fails in general. In fact, it fails for condensed abelian groups, which is a mayor annoyance.

Projective condensed abelian groups $M$ are in fact exactly the retracts of $\mathbb Z[S]$ where $S$ is an extremally disconnected profinite set, or even the Stone-Cech compactification of a discrete set. (A curious, slightly nontrivial fact is that in that case also $M\oplus \mathbb Z[S]\cong \mathbb Z[S]$.) In any case, the question becomes whether $\mathbb Z[S]\otimes \mathbb Z[T]\cong \mathbb Z[S\times T]$ is projective. This can be shown never to be the case when $S$ and $T$ are infinite, for example via Proposition 4.8 in the Analytic Geometry lectures.

I'd be extremely interested in the answer to the following question:

If $S$ and $T$ are extremally disconnected profinite sets, does $\mathbb Z[S\times T]$ have finite projective dimension?

I've spent quite a bit of time in this, but couldn't make progress in either direction. In some ways, it seems unlikely, but a positive answer would be really helpful.

This is nit an answer, but it's too long for a comment.

1- Note that many spaces don't actually have nonzero projective sheaves, and in fact, the latter are quite rare; so this question may not be so relevant.

2- Also, another point is that it seems unlikely to be true : the natural "unbiased" version of this statement ("a finite tensor product of projective sheaves is projective") fails - the empty tensor product is the unit, which is the constant $\mathbb Z$ sheaf, and it's certainly not projective in general (it would mean that epimorphisms of sheaves are epimorphisms of presheaves, which is rarely the case). Of course this doesn't prove that the statement is wrong because the empty tensor product is somewhat special; but still it makes it seem unlikely.

3- The internal hom argument will also likely not work in general : indeed, try to see what it means for $\mathcal{Hom}(G,-)$ to preserve epimorphisms. Start from an epimorphism $F\to H$, then this says that for any $U$ and any map $G_{\mid U} \to H_{\mid U}$, there is locally a lift. Now, of course, $F\to H$ is an epimorphism no matter where you restrict it, but to get a lift, you "morally" will need the resrictions of $G$ to still be projective, which is related to $i_*$ (where $i : U\to X$ is the inclusion) preserving epimorphisms which, in general, it doesn't.

As you can see, neither of 1,2 or 3 shows that the statement is wrong, so again, this does not answer your question, but it still makes it seem like the answer will be no in general. However, it will be true over stupid examples such as discrete spaces; and in fact, the rareness of projective sheaves will likely make it hard to actually find a counterexample if the statement is false.

It could even be (I wouldn't count on it) that spaces that are weird enough to have projective sheaves are also weird enough for this type of thing to actually hold