Verified answer

[tex]\displaystyle\bf\\1)\\\\Cos^{2} \Big(\pi -\alpha \Big) tg\Big(\pi +\alpha \Big) tg\Big(\frac{3\pi }{2} -\alpha \Big)+Sin\Big(2\pi -\alpha \Big) Cos\Big(\frac{\pi }{2} +\alpha \Big)=\\\\\\=Cos^{2} \alpha \cdot \underbrace{tg\alpha \cdot Ctg\alpha}_{1} -Sin\alpha \cdot \Big(-Sin\alpha \Big)=Cos^{2} \alpha +Sin^{2} \alpha =1\\\\\\2)[/tex]

[tex]\displaystyle\bf\\10Ctg\frac{3\pi }{4} Sin\frac{5\pi }{4} Cos\frac{7\pi }{4} =10Ctg\Big(\pi -\frac{\pi }{4} \Big) Sin\Big(\pi +\frac{\pi }{4} \Big) Cos\Big(2\pi -\frac{\pi }{4}\Big)=\\\\\\=10\cdot\Big(-Ctg\frac{\pi }{4}\Big)\cdot\Big(-Sin\frac{\pi }{4} \Big)\cdot Cos\frac{\pi }{4}=10\cdot 1\cdot \frac{\sqrt{2} }{2} \cdot\frac{\sqrt{2} }{2} =5\\\\\\3)\\\\\frac{Sin^{2}(\pi -t) }{1+Sin(\frac{3\pi }{2} +t)} -Cos(2\pi -t)=\frac{Sin^{2}t }{1-Cost} -Cost=[/tex]

[tex]\displaystyle\bf\\=\frac{Sin^{2} t-Cost+Cos^{2} t}{1-Cost} =\frac{1-Cost}{1-Cost}=1[/tex]