2. Given in $\triangle A B C$, $B C=5, A C=4, \cos (A-B)=\frac{7}{8}$, then $\cos C=$
2. $\frac{11}{16}$.

Since $B C>A C$, hence $\angle B A C>\angle B$. Take a point $D$ on side $B C$ such that $\angle B A D=\angle B$, then $\angle D A C=$ $\angle B A C-\angle B$. Let $A D=B D=x$, then $D C=5-x$. In $\triangle A D C$, by the cosine rule, we get $x=3$. Again by the cosine rule, we get $\cos C=\frac{11}{16}$.