Đáp án:

a) 
$Sin(\alpha -\frac{\pi}{6})=\frac{-\sqrt{3}}{2}\\
cos(\alpha +\frac{\pi}{4})=\frac{\sqrt{6}+\sqrt{2}}{4}$
b)
$sin(\alpha +\frac{\pi}{4} )=\frac{3\sqrt{34}}{34}\\
tan(\alpha -\frac{\pi}{4})=\frac{5}{3}$

Giải thích các bước giải:

a) Ta có $sin^2\alpha +cos^2\alpha =1\\
\Rightarrow cos^2\alpha =1-sin^2\alpha =1-(\frac{-1}{2})^2=\frac{3}{4}\\
\Rightarrow cos\alpha =\pm \sqrt{\frac{3}{4}}=\pm \frac{\sqrt{3}}{2}$
Vì $\frac{3\pi}{2}<\alpha <2\pi \Rightarrow cos=\frac{\sqrt{3}}{2}\\
Sin(\alpha -\frac{\pi}{6})=sinxcos\frac{\pi}{6}-cosxsin\frac{\pi}{6}\\
=(\frac{-1}{2}).\frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{2}.\frac{1}{2}\\
=\frac{-\sqrt{3}}{2}\\
cos(\alpha +\frac{\pi}{4})=cos\alpha cos\frac{\pi}{4}-sin\alpha sin\frac{\pi}{4}\\
=\frac{\sqrt{3}}{2}.\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}.\frac{-1}{2}\\
=\frac{\sqrt{6}+\sqrt{2}}{4}$
b)
Ta có $1+tan^2\alpha =\frac{1}{cos^2\alpha }\\
\Rightarrow cos^2\alpha =\frac{1}{1+tan^2\alpha }=\frac{1}{1+(-4)^2}=\frac{1}{17}\\
\Rightarrow cos\alpha =\pm \frac{\sqrt{17}}{17}$
Vì $\frac{\pi}{2}<\alpha <\pi \Rightarrow cos=\frac{-\sqrt{17}}{17}$
Mặt khác $tan\alpha =\frac{sin\alpha }{cos\alpha }\\
\Rightarrow sin\alpha =tan\alpha .cos\alpha =(-4).\frac{-\sqrt{17}}{17}=\frac{4\sqrt{17}}{17}\\
sin(\alpha +\frac{\pi}{4} )=sin\alpha cos\frac{\pi}{4}+cos\alpha sin\frac{\pi}{4}\\
=\frac{4\sqrt{17}}{17}.\frac{\sqrt{2}}{2}+\frac{-\sqrt{17}}{17}.\frac{\sqrt{2}}{2}\\
=\frac{3\sqrt{34}}{34}\\
tan(\alpha -\frac{\pi}{4})=\frac{tan\alpha -tan\frac{\pi}{4}}{1+tan\alpha tan\frac{\pi}{4}}\\
=\frac{-4-1}{1+(-4).1}=\frac{5}{3}$