已知 $\tan \left(\frac{\pi}{4}+\alpha\right)=\frac{1}{2}$.

(I) 求 $\tan \alpha$ 的值;

(II) 求 $\frac{\sin (2 \alpha+2 \pi)-\sin ^{2}\left(\frac{\pi}{2}-\alpha\right)}{1-\cos (\pi-2 \alpha)+\sin ^{2} \alpha}$ 的值.
(I) $\tan \left(\frac{\pi}{4}+\alpha\right)=\frac{\tan \frac{\pi}{4}+\tan \alpha}{1-\tan \frac{\pi}{4} \tan \alpha}=\frac{1+\tan \alpha}{1-\tan \alpha}=\frac{1}{2}$,解得 $\tan \alpha=-\frac{1}{3}$;

(II) $\frac{\sin (2 \alpha+2 \pi)-\sin ^{2}\left(\frac{\pi}{2}-\alpha\right)}{1-\cos (\pi-2 \alpha)+\sin ^{2} \alpha}=\frac{\sin 2 \alpha-\cos ^{2} \alpha}{1+\cos 2 \alpha+\sin ^{2} \alpha}$

$=\frac{2 \sin \alpha \cos \alpha-\cos ^{2} \alpha}{2 \cos ^{2} \alpha+\sin ^{2} \alpha}=\frac{2 \tan \alpha-1}{2+\tan ^{2} \alpha}=-\frac{15}{19}$.