2. Given $\sin ^{2} \alpha+\cos ^{2} \beta-\sin \alpha \cdot \cos \beta=\frac{3}{4}$, $\sin (\alpha-\beta) \neq \frac{1}{2}$. Then $\sin (\alpha+\beta)=$ $\qquad$ .
$$
\begin{array}{l}
\text { 2. } \frac{1}{2} \text {. } \\
0=\sin ^{2} \alpha+\cos ^{2} \beta-\sin \alpha \cdot \cos \beta-\frac{3}{4} \\
=\frac{1-\cos 2 \alpha}{2}+\frac{1+\cos 2 \beta}{2}- \\
\frac{1}{2}[\sin (\alpha+\beta)+\sin (\alpha-\beta)]-\frac{3}{4} \\
=\frac{1}{4}-\frac{1}{2}(\cos 2 \alpha-\cos 2 \beta)- \\
\frac{1}{2}[\sin (\alpha+\beta)+\sin (\alpha-\beta)] \\
=\frac{1}{4}+\sin (\alpha+\beta) \cdot \sin (\alpha-\beta)- \\
\frac{1}{2}[\sin (\alpha+\beta)+\sin (\alpha-\beta)] \\
=\left[\frac{1}{2}-\sin (\alpha+\beta)\right]\left[\frac{1}{2}-\sin (\alpha-\beta)\right] \text {. } \\
\end{array}
$$

Since $\sin (\alpha-\beta) \neq \frac{1}{2}$, then,
$$
\sin (\alpha+\beta)=\frac{1}{2} .
$$