Verified answer

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

[tex]\displaystyle\bf\\3)\\\\\frac{tg(\pi -\beta ) Cos(\pi -\beta ) tg\Big(\frac{\pi }{2} -\beta \Big)}{Sin\Big(\frac{\pi }{2}-\beta \Big) Ctg\Big(\frac{\pi }{2} +\alpha \Big) tg\Big(\frac{3\pi }{2} +\alpha\Big) } =\frac{-tg\beta\cdot (-Cos\beta )\cdot Ctg\beta }{Cos\beta \cdot (-tg\alpha )\cdot (-Ctg\alpha )} =\\\\\\=\frac{Cos\beta }{Cos\beta } =1\\\\\\4)[/tex]

[tex]\displaystyle\bf\\\frac{tg\Big(\frac{3\pi }{2} +\alpha \Big) Sin\frac{3\pi }{2} Sin\frac{16\pi }{9} Cos\frac{13\pi }{18} }{Ctg(\pi -\alpha ) Cos\frac{5\pi }{18} Sin\frac{11\pi }{9}Cos2\pi } =\\\\\\=\frac{-Ctg\alpha \cdot (-1)\cdot Sin(2\pi -\frac{2\pi }{9} )\cdot Cos(\pi -\frac{5\pi }{18} ) }{-tg\alpha\cdot Cos\frac{5\pi }{18} \cdot Sin(\pi +\frac{2\pi }{9} )\cdot 1 } =[/tex]

[tex]\displaystyle\bf\\=\frac{Ctg\alpha \cdot (-Sin\frac{2\pi }{9} )\cdot Cos\frac{5\pi }{18} }{-tg\alpha \cdot Cos\frac{5\pi }{18} \cdot(-Sin\frac{2\pi }{9} )} =-\frac{Ctg\alpha }{tg\alpha } =-\frac{Ctg\alpha }{\frac{1}{Ctg\alpha } } =-Ctg^{2}\alpha[/tex]