a,
$\dfrac{1}{\cos^2\alpha}=1+\tan^2\alpha$
$\Rightarrow \cos\alpha=\sqrt{\dfrac{1}{1+(\frac{3}{4})^2}}=0,8$
$\Rightarrow \sin\alpha=\sqrt{1-\cos^2\alpha}=0,6$
b,
$\dfrac{1}{\sin^2\alpha}=1+\cot^2\alpha$
$\Rightarrow \sin\alpha=\sqrt{\dfrac{1}{1+(\frac{5}{12})^2}}=\dfrac{12}{13}$
$\Rightarrow \cos\alpha=\sqrt{1-\sin^2\alpha}=\dfrac{5}{13}$