2. Given $\sin \alpha+\cos \alpha=\frac{\sqrt{2}}{2}$. Then $\sin ^{4} \alpha+\cos ^{4} \alpha=$ $\qquad$
2. $\frac{7}{8}$.

Squaring the given equation yields
$$
\begin{array}{l}
1+2 \sin \alpha \cdot \cos \alpha=\frac{1}{2} \\
\Rightarrow \sin \alpha \cdot \cos \alpha=-\frac{1}{4} . \\
\text { Therefore, } \sin ^{4} \alpha+\cos ^{4} \alpha \\
=\left(\sin ^{2} \alpha+\cos ^{2} \alpha\right)^{2}-2(\sin \alpha \cdot \cos \alpha)^{2}=\frac{7}{8} .
\end{array}
$$