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1) $sinθ=\frac{1}{6}$

and  $tanθ=\frac{sinθ}{cosθ}$ <=> $cosθ=\frac{sinθ}{tanθ}=\frac{\frac{1}{6}}{\frac{-\sqrt{35}}{35}}=\frac{-\sqrt{35}}{6 }$ 

so $cosθ= \frac{-\sqrt{35}}{6}$

2) $cosθ=cos(-θ)=\frac{\sqrt{3}}{4}$

and $cos^2θ + sin^2θ =1$ 

so  $sinθ =\sqrt{1-cos^2θ} = \sqrt{\frac{16}{16}-\frac{3}{16}}=\sqrt{\frac{13}{16}}=\frac{\sqrt{13}}{-4}$

so $sinθ=-\frac{\sqrt{13}}{4}$

Hope this helps!