13. Given $\sin \alpha+\sin \beta+\sin \gamma=0, \cos \alpha+\cos \beta+\cos \gamma=0$, find $\cos ^{2} \alpha+\cos ^{2} \beta+$ $\cos ^{2} \gamma$.

13. From the conditions, we know that $\sin \alpha+\sin \beta=-\sin \gamma, \cos \alpha+\cos \beta=-\cos \gamma$. Squaring and adding these two equations, we get
$$
\begin{array}{l}
2+2 \cos (\alpha-\beta)=1 \text {, so } 2 \cos (\alpha-\beta)=-1, \cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} \gamma=\cos ^{2} \alpha+\cos ^{2} \beta+(\cos \alpha+ \\
\cos \beta)^{2}=2+(\cos 2 \alpha+\cos 2 \beta)+2 \cos \alpha \cos \beta=2+2 \cos (\alpha+\beta) \cdot \cos (\alpha-\beta)+2 \cos \alpha \cos \beta= \\
2-\cos (\alpha+\beta)+2 \cos \alpha \cos \beta=2+\cos (\alpha-\beta)=\frac{3}{2} .
\end{array}
$$