8. Given $\sin \alpha+\sin \beta=\sin \gamma$, $\cos \alpha+\cos \beta=-\cos \gamma$.
Then $\cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} \gamma=$ $\qquad$
8. $\frac{3}{2}$.

Squaring the given condition, we get
$$
\begin{array}{l}
\sin ^{2} \alpha+\sin ^{2} \beta+2 \sin \alpha \cdot \sin \beta=\sin ^{2} \gamma, \\
\cos ^{2} \alpha+\cos ^{2} \beta+2 \cos \alpha \cdot \cos \beta=\cos ^{2} \gamma .
\end{array}
$$

Adding the two equations yields $\cos (\alpha-\beta)=-\frac{1}{2}$;
Subtracting the two equations yields
$$
\begin{array}{l}
\cos 2 \gamma=\cos 2 \alpha+\cos 2 \beta+2 \cos (\alpha+\beta) \\
=2 \cos (\alpha+\beta) \cdot \cos (\alpha-\beta)+2 \cos (\alpha+\beta) \\
=\cos (\alpha+\beta) .
\end{array}
$$

Therefore, $\cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} \gamma$
$$
\begin{array}{l}
=\frac{3}{2}+\frac{1}{2}(\cos 2 \alpha+\cos 2 \beta+\cos 2 \gamma) \\
=\frac{3}{2}+\frac{1}{2}[2 \cos (\alpha-\beta) \cdot \cos (\alpha+\beta)+\cos 2 \gamma] \\
=\frac{3}{2}+\frac{1}{2}[-\cos (\alpha+\beta)+\cos 2 \gamma]=\frac{3}{2} .
\end{array}
$$