Помогите пожалуйста !!!!

1) α - угол третьей четверти, значит Sinβ < 0

$Sin\beta=-\sqrt{1-Cos^{2}\beta}=-\sqrt{1-(-\frac{7}{25})^{2} }=-\sqrt{1-\frac{49}{625} }=-\sqrt{\frac{576}{625} }=- \frac{24}{25}$

$2)\frac{3Cos\alpha+10Sin\alpha}{Cos\alpha }=\frac{\frac{3Cos\alpha }{Cos\alpha }+\frac{10Sin\alpha }{Cos\alpha}}{\frac{Cos\alpha }{Cos\alpha}}=\frac{3+10tg\alpha }{1}=3+10*4=43$

$3)tg\alpha+Ctg\alpha=9\\\\(tg\alpha+Ctg\alpha)^{2} =9^{2}\\\\tg^{2}\alpha+2tg\alpha Ctg\alpha+Ctg^{2}\alpha=81\\\\tg^{2}\alpha+2+Ctg^{2}\alpha=81\\\\tg^{2}\alpha+Ctg^{2}\alpha=79$

$4)\frac{(Sin(90^{0}+3\alpha)+Cos(180^{0}+\alpha))*(Sin(180^{0}-3\alpha)-Cos(270^{0}+\alpha))}{1+Cos(180^{0}-2\alpha)}=\frac{(Cos3\alpha-Cos\alpha)(Sin3\alpha-Sin\alpha)}{1-Cos2\alpha }=\frac{-2Sin\frac{3\alpha+\alpha}{2}Sin\frac{3\alpha-\alpha}{2}*2Sin\frac{3\alpha-\alpha}{2}Cos\frac{3\alpha+\alpha}{2}}{2Sin^{2}\alpha} =\frac{-2Sin2\alpha Sin\alpha Sin\alpha Cos2\alpha}{Sin^{2}\alpha}=-2Sin2\alpha Cos2\alpha =-Sin4\alpha\\\\\alpha=7,5^{0} \\\\-Sin(4*7,5^{0})=-Sin30^{0}=-\frac{1}{2}$