Срочно!!! Решите номер 2,4,5 (2 и 4 ОБЯЗАТЕЛЬНО 5если сможете)

Решите задачу:

$8Sin2 \beta Cos2 \beta Cos4 \beta =4*(2Sin2 \beta Cos2 \beta )Cos4 \beta =4Sin4 \beta Cos4 \beta =$ $2*(2Sin4 \beta Cos4 \beta) = 2Sin8 \beta \\\\ \frac{Sin(2 \alpha +3 \beta )-Sin(2 \alpha -3 \beta )}{Cos(2 \alpha -3 \beta )-Cos(2 \alpha +3 \beta )}= \frac{Sin2 \alpha Cos3 \beta +Cos2 \alpha Sin3 \beta -Sin2 \alpha Cos3 \beta +Cos2 \alpha Sin3 \beta }{Cos2 \alpha Cos3 \beta +Sin2 \alpha Sin3 \beta -Cos2 \alpha Cos3 \beta +Sin2 \alpha Sin3 \beta }$ $= \frac{2Cos2 \alpha Sin3 \beta }{2Sin2 \alpha Sin3 \beta } =Ctg2 \alpha$

$tgx = \frac{1}{2} \\\\tg ^{2} x+1= \frac{1}{Cos ^{2}x } \\\\Cos ^{2}x= \frac{1}{tg ^{2}x+1 }\\\\Cos2x=2Cos ^{2}x-1=2* \frac{1}{( \frac{1}{2}) ^{2} +1 }-1= 2* \frac{4}{5} -1= \frac{3}{5} \\\\Cos4x=2Cos ^{2}2x-1 =[tex]2*( \frac{3}{5}) ^{2}-1=2* \frac{9}{25}-1= \frac{18}{25}-1=- \frac{7}{25}\\\\Sin4x= \sqrt{1-Cos ^{2}4x } = \sqrt{1- \frac{49}{625} }= \sqrt{ \frac{576}{625} } = \frac{24}{25}$