Let $\otimes$ denotes the usual tensor products of complexes and symbols live in the category of chain complexes of $R$-modules. Let $X$ be a dg-flat complex (i.e. $X\_n$ is flat for each n and $X\otimes E$ is exact for each exact complex E). We can define a new complex $X^+={\rm Hom}\_Z(X, \mathbb{Q}/\mathbb{Z})$. Is it true to say that $X^+$ is a dg-injective chain complex?



dg-flat complexes and their characters
I take 'dg-injective' as fibrant in the injective model structure on complexes of right dg-$R$-modules, whose weak equivalences are the quasi-isomorphisms and whose cofibrations are the levelwise injections ($X$ is therefore a complex of left $R$-modules). Then the answer is yes. I'll prove it satisfies the required lifting property. Consider a commutative square of complexes of right dg-$R$-modules
$$\begin{array}{ccc}
A&\rightarrow &\hom\_{\mathbb Z}(X,\mathbb Q/\mathbb Z)\\
\downarrow&&\downarrow\\
B&\rightarrow&0
\end{array}
$$
where $i\colon A\stackrel{\sim}\rightarrowtail B$ is a quasi-isomorphism which is also a levelwise inclusion. The adjoint of this diagram is a commutative square of complexes of abelian groups
$$\begin{array}{ccc}
A\otimes\_R X&\rightarrow &\mathbb Q/\mathbb Z\\
\downarrow&&\downarrow\\
B\otimes\_R X&\rightarrow&0
\end{array}
$$
Here $i\otimes\_R X\colon A\otimes\_R X\rightarrow B\otimes\_R X$ is levelwise injective since $X$ is levelwise flat and direct sums in the category of abelian groups are exact. Moreover, the cokernel of $i\otimes\_RX$ is $(B/A)\otimes\_R X$, which is exact since $B/A$ is exact (recall that $i$ is a an injective quasi-isomorphism). Therefore $i\otimes\_RX$ is a cofibration in the injective model structure on chain complexes of abelian groups. Hence the latter square has a lifting $B\otimes\_RX\rightarrow \mathbb Q/\mathbb Z$. The adjoint map $B\rightarrow \hom\_{\mathbb Z}(X,\mathbb Q/\mathbb Z)$ is a lifting of the former. This concludes the argument.


