1) Дано cos a= -4/5 и угол во II четверти

[tex]\displaystyle sin^2 a=1-cos^2a=1-(-\frac{4}{5})^2=1-\frac{16}{25}=\frac{9}{25}}\\\\sina=\pm \frac{3}{5}[/tex]

sin a во II четверти положительный. Значит sin a= 3/5

[tex]\displaystyle tga=\frac{sina}{cosa}=\frac{3/5}{-4/5}=\frac{3}{5}*\frac{5}{-4}=-\frac{3}{4}[/tex]

[tex]\displaystyle cos 2a=cos^2a-sin^2a=1-2sin^2a=1-2*(\frac{3}{5})^2=1-2*\frac{9}{25}=\\\\=\frac{25-18}{25}=\frac{7}{25}[/tex]

2)

[tex]\displaystyle cos135`=cos(90+45)=-sin 45=-\frac{\sqrt{2}}{2}\\\\sin \frac{8\pi }{3}=sin(2\pi +\frac{2\pi }{3})=sin\frac{2\pi }{3}=\frac{\sqrt{3}}{2} \\\\tg \frac{7\pi }{3}=tg(2\pi +\frac{\pi }{3})=tg\frac{\pi }{3}=\sqrt{3}\\\\ cos^2\frac{\pi }{8}-sin^2\frac{\pi }{8}=cos(2*\frac{\pi }{8})==cos\frac{\pi }{4}=\frac{\sqrt{2}}{2}[/tex]