Решите с объяснением, пожалуйста)

Решите задачу:

$1)2Sin120^{o} -Ctg240^{o} +3Cos330^{o} =2Sin(180^{o}-60^{o})-Ctg(180^{o}+60^{o})+3Cos(360^{o}-30^{o} )=2Sin60^{o}-Ctg60^{0}+3Cos30^{o}=2*\frac{\sqrt{3} }{2}-\frac{\sqrt{3} }{3}+3*\frac{\sqrt{3} }{2}=\frac{6\sqrt{3} -2\sqrt{3}+9\sqrt{3}}{6} =\frac{13\sqrt{3} }{6}$

$2)Ctg105^{o}=Ctg(60^{o}+45^{o})=\frac{Ctg60^{o}*Ctg45^{o} -1 } {Ctg60^{o}+Ctg45^{o}}=\frac{\frac{\sqrt{3} }{3}*1-1 }{\frac{\sqrt{3} }{3}+1 } =\frac{(\sqrt{3}-3)*3 }{(\sqrt{3}+3)*3 }=\frac{\sqrt{3}(1-\sqrt{3} )}{\sqrt{3}(1+\sqrt{3})}=\frac{1-\sqrt{3} }{1+\sqrt{3}}$

$3)Cos\frac{11\pi }{10}Cos\frac{\pi }{10}+Sin\frac{11\pi }{10}Sin\frac{\pi }{10}=Cos(\frac{11\pi }{10}-\frac{\pi }{10})= Cos\pi =-1$

$4)\frac{tg24^{o} +tg6^{o} }{1-tg24^{0}tg6^{o}}=tg(24^{o}+6^{o})=tg30^{o} =\frac{\sqrt{3} }{3}$

$5)\frac{Cos^{2}\frac{\pi }{5}+Sin^{2}\frac{\pi }{5}}{2Sin\frac{\pi }{8}Cos\frac{\pi }{8}}=\frac{1}{Sin\frac{\pi }{4} }=\frac{1}{\frac{1}{\sqrt{2} } }=\sqrt{2}$