若 $\alpha \in(\pi, 2 \pi)$ ，且 $\sin \alpha+\cos \alpha=\frac{\sqrt{2}}{4}$.

(1)求 $\cos ^{2} \alpha-\cos ^{4} \alpha$ 的值;

(2)求 $\sin \alpha-\cos \alpha$ 的值.
)因为 $\sin \alpha+\cos \alpha=\frac{\sqrt{2}}{4}$ ，所以 $(\sin \alpha+\cos \alpha)^{2}=\left(\frac{\sqrt{2}}{4}\right)^{2}$, 即 $1+2 \sin \alpha \cos \alpha=\frac{1}{\frac{1}{8}}$, 所以 $\sin \alpha \cos \alpha==^{\frac{7}{16}}$, 所以 $\cos ^{2} \alpha-\cos ^{4} \alpha=\cos ^{2} \alpha\left(1-\cos ^{2} \alpha\right)=\cos ^{2} \alpha \sin ^{2} \alpha=(\sin \alpha \cos \alpha)^{2}=\left(-\frac{7}{16}\right)^{2} =\frac{49}{256}$.

(2) $(\sin \alpha-\cos \alpha)^{2}=1-2 \sin \alpha \cos \alpha=1-2 x^{\left(-\frac{7}{16}\right)}=\frac{15}{8}$ ，由(1)知 $\sin \alpha \cos \alpha=-\frac{7}{16}<0$ ，又 $\alpha \in(\pi, 2 \pi)$,所以 $\alpha \in^{\left(\frac{3 \pi}{2}, 2 \pi\right)}$,所以 $\sin \alpha<0, \cos \alpha>0$, 所以 $\sin \alpha-\cos \alpha<0$, 所以 $\sin \alpha-\cos \alpha=-\frac{\sqrt{30}}{4}$.