Срочно помогите по тригонометрий упростите

Решите задачу:

$1) sin^{4} \frac{23 \pi }{12}- cos^{4} \frac{13 \pi }{12}=sin^4(2 \pi - \frac{ \pi}{12})-cos^4( \pi + \frac{ \pi }{12})=sin^4 \frac{ \pi }{12}-cos^4 \frac{ \pi }{12} \\ =(sin^2 \frac{ \pi }{12}-cos^2 \frac{ \pi }{12})(sin^2 \frac{ \pi }{12}+cos^2 \frac{ \pi }{12})=-cos(2* \frac{ \alpha }{12})=-cos \frac{ \pi }{6}=- \frac{ \sqrt{3} }{2};$
$2) tg \frac{7 \pi }{8}+ctg \frac{7 \pi}{8}=tg( \pi - \frac{ \pi }{8})+ctg( \pi - \frac{ \pi }{8})=-tg \frac{ \pi }{8}-ctg \frac{ \pi }{8}= \\=-\frac{sin \frac{ \pi }{4} }{1+cos \frac{ \pi }{4}}- \frac{1+cos \frac{ \pi }{4}}{sin \frac{ \pi }{4}}=-( \frac{ \frac{ \sqrt{2} }{2} }{1+ \frac{ \sqrt{2} }{2}}+ \frac{1+ \frac{ \sqrt{2} }{2} }{ \frac{ \sqrt{2} }{2}})=-( \frac{ \frac{1}{2}+ \frac{3}{2}+ \sqrt{2}}{ \frac{ \sqrt{2} }{2}+ \frac{1}{2}})=$
$=- \frac{2(2+ \sqrt{2} )}{1+ \sqrt{2}}=- \frac{2(2+ \sqrt{2})(1- \sqrt{2})}{1-2}=-2 \sqrt{2};$
$3) ctg \frac{5 \pi }{8}+ctg \frac{9 \pi }{8}= \frac{sin \frac{7 \pi }{4}}{sin \frac{5 \pi }{8}sin \frac{9 \pi }{8}}= \frac{- \frac{ \sqrt{2}}{2}}{ \frac{1}{2}(cos \frac{ \pi }{2}-cos \frac{7 \pi }{4})}= \\ =\frac{- \frac{ \sqrt{2}}{2}}{ \frac{1}{2}(0- \frac{ \sqrt{2}}{2})}= \frac{ \sqrt{2} }{2}: \frac{ \sqrt{2} }{4}=2;$