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$1)\frac{Sin^{2}2\alpha}{Sin(\alpha+\frac{3\pi }{2})*Cos(\frac{\pi }{2}-\alpha)}=\frac{Sin^{2}2\alpha}{-Cos\alpha*Sin\alpha}=-\frac{4Sin^{2}\alpha Cos^{2}\alpha}{Cos\alpha Sin\alpha}=-4Sin\alpha Cos\alpha$

$2)Sin\alpha +Cos\alpha=\sqrt{2}\\\\(Sin\alpha+Cos\alpha)^{2}=(\sqrt{2})^{2} \\\\Sin^{2}\alpha+2Sin\alpha Cos\alpha +Cos^{2}\alpha=2\\\\1+2Sin\alpha Cos\alpha=2\\\\2Sin\alpha Cos\alpha=1\\\\3)-4Sin\alpha Cos\alpha=2Sin\alpha Cos\alpha*(-2)=1*(-2)=-2$

Ответ : - 2