$△DEF$ is right-angled at $E$. If $\cot D=\frac{8}{15}$, then what is the value of $\cot F$?

In a right-angled triangle $△DEF$ at $E$, the two acute angles $D$ and $F$ are complementary:
$$D+F=90^{\circ }⟹F=90^{\circ }-D$$
Using trigonometric complementary identities:
$$\cot F=\cot (90^{\circ }-D)=\tan D$$
Since $\cot D=\frac{8}{15}$ and $\tan D=\frac{1}{\cot D}$:
$$\tan D=\frac{1}{8/15}=\frac{15}{8}$$
Therefore, $\cot F=\frac{15}{8}$.