(14) Given $\sin \alpha+\sin \beta=\frac{\sqrt{6}}{3}, \cos \alpha+\cos \beta=\frac{\sqrt{3}}{3}$, then $\cos ^{2} \frac{\alpha-\beta}{2}=$
(14) $\frac{1}{4}$ Hint:
$$
\begin{aligned}
& (\sin \alpha+\sin \beta)^{2}=\frac{2}{3},(\cos \alpha+\cos \beta)^{2}=\frac{1}{3} \\
\Rightarrow & \sin ^{2} \alpha+\sin ^{2} \beta+2 \sin \alpha \sin \beta=\frac{2}{3}, \\
& \cos ^{2} \alpha+\cos ^{2} \beta+2 \cos \alpha \cos \beta=\frac{1}{3} \\
\Rightarrow & 2+2 \sin \alpha \sin \beta+2 \cos \alpha \cos \beta=1 \\
\Rightarrow & 1+\cos (\alpha-\beta)=\frac{1}{2} \\
\Rightarrow & \cos ^{2} \frac{\alpha-\beta}{2}=\frac{1}{4} .
\end{aligned}
$$