1. Let $\sin x+\cos x$ $=\frac{1}{2}$, then $\sin ^{3} x+\cos ^{3} x=$ $\qquad$
Since $(\sin x+\cos x)^{2}=1+2 \sin x \cos x=\frac{1}{4} \Rightarrow \sin x \cos x=-\frac{3}{8}$,
thus $\sin ^{3} x+\cos ^{3} x=(\sin x+\cos x)(1-\sin x \cos x)=\frac{1}{2} \cdot \frac{11}{8}=\frac{11}{16}$.