Упростить второй пример. Напишите решение!

Решите задачу:

$1)\; \; \; tg \alpha =-\sqrt5\\\\1+tg^2 \alpha =\frac{1}{cos^2 \alpha }\; \; \Rightarrow \; \; 1+(-\sqrt5)^2=1+5=6=\frac{1}{cos^2 \alpha }\\\\cos^2 \alpha =\frac{1}{6}\\\\ \frac{5cos2 \alpha +3}{3-8cos^2 \alpha } =[\, cos2 \alpha =2cos^2 \alpha -1\, ]= \frac{5(2cos^2 \alpha -1)+3}{3-8cos^2 \alpha } =\frac{10cos^2 \alpha -2}{3-8cos^2 \alpha }=\\\\= \frac{\frac{10}{6}-2}{3-\frac{8}{6}} =\frac{10-12}{18-8}=\frac{-2}{10}=-0,2$

$2)\; \; ctg \alpha =-2\\\\1+ctg^2 \alpha =\frac{1}{sin^2 \alpha }\; \; \Rightarrow \; \; 1+(-2)^2=1+4=5=\frac{1}{sin^2 \alpha }\\\\sin^2 \alpha =\frac{1}{5}\; \; \to \; \; sin \alpha =\pm \frac{1}{\sqrt5}\; ,\\\\cos^2 \alpha =1-sin^2 \alpha =1-\frac{1}{5}=\frac{4}{5}\; ,\; \; cos \alpha =\mp\, \frac{2}{\sqrt5}\\\\ 1)\; \frac{2-4sin^2 \alpha }{3+sin2 \alpha } = \frac{2-4\cdot \frac{1}{5}}{3+2sin \alpha \cdot cos \alpha }= \frac{\frac{6}{5}}{3+2\cdot \frac{1}{\sqrt5}\cdot (-\frac{2}{\sqrt5})}=$

$=\frac{\frac{6}{5}}{3-\frac{4}{5}}=\frac{6}{15-4}=\frac{6}{11}\\\\b)\; \frac{2-4sin^2 \alpha }{3+2sin2 \alpha }= \frac{2-\frac{4}{5}}{3+2\cdot (-\frac{1}{\sqrt5})\cdot \frac{2}{\sqrt5}} =\frac{6}{11}$