7. In $\triangle A B C$, it is known that
$$
B C=5, A C=4, \cos (A-B)=\frac{7}{8} \text {. }
$$

Then $\cos C=$
7. $\frac{11}{16}$.
From $B C>A C$, we get $\angle A>\angle B$.
As shown in Figure 3, construct $A D$

such that $\angle B A D=\angle B$. Then
$\angle D A C$
$=\angle A-\angle B$.
Let $A D=B D=x$.
Then $D C=5-x$.

In $\triangle A D C$, by the Law of Cosines, we get $x=3$.
Again, by the Law of Cosines, we get $\cos C=\frac{11}{16}$.