8. Let $I$ be the incenter of $\triangle A B C$, and
$$
3 \overrightarrow{I A}+4 \overrightarrow{I B}+5 \overrightarrow{I C}=\mathbf{0} \text {. }
$$

Then the size of $\angle C$ is $\qquad$
8. $90^{\circ}$.

As shown in Figure 2, let $a$, $b$, and $c$ be the sides opposite to $\angle A$, $\angle B$, and $\angle C$ respectively. Draw lines through point $I$ parallel to the three sides of $\triangle ABC$, intersecting at points $D$, $N$, $E$, $M$, $F$, and $G$.

Since $I$ is the incenter of $\triangle ABC$, we have
$$
\begin{array}{l}
A N=A G=\lambda_{1}, B D=B E=\lambda_{2}, \\
C F=C M=\lambda_{3}, \\
\overrightarrow{A I}=\overrightarrow{A N}+\overrightarrow{A G}=\lambda_{1}\left(\frac{\overrightarrow{A B}}{c}+\frac{\overrightarrow{A C}}{b}\right), \\
\overrightarrow{B I}=\overrightarrow{B D}+\overrightarrow{B E}=\lambda_{2}\left(\frac{\overrightarrow{B A}}{c}+\frac{\overrightarrow{B C}}{a}\right), \\
\overrightarrow{C I}=\overrightarrow{C F}+\overrightarrow{C M}=\lambda_{3}\left(\frac{\overrightarrow{C A}}{b}+\frac{\overrightarrow{C B}}{a}\right). \\
\text { Therefore, } \frac{\overrightarrow{A I}}{\lambda_{1}}+\frac{\overrightarrow{B I}}{\lambda_{2}}+\frac{\overrightarrow{C I}}{\lambda_{3}}=\mathbf{0}. \\
\text { Also, } M N / / A C, \text { so } \frac{A N}{c}=\frac{C M}{a}.
\end{array}
$$

Similarly, $\frac{B D}{c}=\frac{C F}{b}$.
Thus, $\frac{\lambda_{1}}{\lambda_{3}}=\frac{c}{a}, \frac{\lambda_{2}}{\lambda_{3}}=\frac{c}{b}$
$\Rightarrow \frac{1}{\lambda_{1}}: \frac{1}{\lambda_{2}}: \frac{1}{\lambda_{3}}=a: b: c$
$$
\Rightarrow a \overrightarrow{I A}+b \overrightarrow{I B}+c \overrightarrow{I C}=\mathbf{0}.
$$

Given $3 \overrightarrow{I A}+4 \overrightarrow{I B}+5 \overrightarrow{I C}=\mathbf{0}$, we find that the ratio of the three sides of $\triangle ABC$ is $a: b: c=3: 4: 5$, which means $\angle C=90^{\circ}$.