8. Given $I$ is the incenter of $\triangle A B C$, and $5 \overrightarrow{I A}=$ $4(\overrightarrow{B I}+\overrightarrow{C I})$. Let $R$ and $r$ be the radii of the circumcircle and incircle of $\triangle A B C$, respectively. If $r=15$, then $R=$ $\qquad$ .
8. 32 .

As shown in Figure 5, take the midpoint $D$ of $B C$. According to the problem, we have
$$
\begin{array}{l}
5 \overrightarrow{I A} \\
=-4(\overrightarrow{I B}+\overrightarrow{I C}) \\
=-8 \overrightarrow{I D} .
\end{array}
$$

Therefore, points $A$, $I$, and $D$ are collinear, and $A B=A C$.
Given $r=I D=15$, we know $I A=24$.
Construct $I E \perp A B$ at point $E$, thus,
$$
\begin{array}{l}
I E=I D=15, \\
\sin \angle B A D=\frac{15}{24}=\frac{5}{8}, \\
\cos \angle B A D=\frac{\sqrt{39}}{8}, \tan \angle B A D=\frac{5}{\sqrt{39}} .
\end{array}
$$

Then, $B C=2 B D=2 A D \tan \angle B A D=10 \sqrt{39}$.
$$
\begin{array}{l}
\text { Also, } \sin \angle B A C=\sin 2 \angle B A D \\
=2 \times \frac{5}{8} \times \frac{\sqrt{39}}{8}=\frac{5 \sqrt{39}}{32},
\end{array}
$$

Thus, $2 R=\frac{B C}{\sin \angle B A C}=64$
$$
\Rightarrow R=32 \text {. }
$$