If there are no boundedness conditions on the complexes then this need not be true. For example, let $F_\bullet=A_\bullet$ be the unbounded complex
$$\cdots\xrightarrow{\times 2}\mathbb{Z}/4\mathbb{Z}\xrightarrow{\times 2}\mathbb{Z}/4\mathbb{Z}\xrightarrow{\times 2}\mathbb{Z}/4\mathbb{Z}\xrightarrow{\times 2}\cdots$$
of $\mathbb{Z}/4\mathbb{Z}$-modules (so it is both exact and a complex of flat $\mathbb{Z}/4\mathbb{Z}$-modules).
Then each component of $F_\bullet\otimes_{\mathbb{Z}/4\mathbb{Z}}A_\bullet$ is isomorphic to
$$\bigoplus_{i\in\mathbb{Z}}\mathbb{Z}/4\mathbb{Z},$$
the kernel of the differential is
$$\bigoplus_{i\in\mathbb{Z}}2\mathbb{Z}/4\mathbb{Z},$$
and the image of the differential is
$$\left\{(a_i)_{i\in\mathbb{Z}}\in\bigoplus_{i\in\mathbb{Z}}2\mathbb{Z}/4\mathbb{Z}
\,\middle\vert\,\sum_{i\in\mathbb{Z}}a_i=0\right\}.$$
So $F_\bullet\otimes_{\mathbb{Z}/4\mathbb{Z}}A_\bullet$ has homology isomorphic to $2\mathbb{Z}/4\mathbb{Z}$ in each degree, and in particular is not exact.
However, with suitable boundedness conditions, it is true.
If $F_\bullet$ is bounded, then an induction argument on the length of $F_\bullet$ shows that $F_\bullet\otimes A_\bullet$ is exact. From this one can prove that $F_\bullet\otimes A_\bullet$ is exact if
(a) either $F_\bullet$ or $A_\bullet$ is bounded, or
(b) both $F_\bullet$ and $A_\bullet$ are bounded above, or
(c) both $F_\bullet$ and $A_\bullet$ are bounded below.
[Alternatively, spectral sequence arguments can be used.]
Similar results hold for the internal Hom, but conditions (b) and (c) are replaced by the condition that one of $F_\bullet$ and $A_\bullet$ is bounded below, and the other is bounded above.
Best Answer
I think this boils down to "given a sequence of Abelian groups, is there a free chain complex having these as homology groups?".
The answer to this is yes. Let $A_n$ be your $n$-th Abelian group. Then it has a free presentation $$0\to F_n'\to F_n\to A_n\to0$$ which is an exact sequence with $F_n$ and $F_n'$ free. Then the homology of the complex $$\cdots\to0\to F_n'\to F_n\to0\to\cdots\tag{*}$$ is just $A_n$ in one position. Now take the direct sum of the (*) over all $n$.