$\dfrac{3\pi}{2}<\alpha<2\pi$
$\to \sin\alpha<0,\cos\alpha>0$
$\cos\alpha=\dfrac{3}{5}$
$\to \sin\alpha=-\sqrt{1-\cos^2\alpha}=-\sqrt{1-\Big(\dfrac{3}{5}\Big)^2}=\dfrac{-4}{5}$
$\tan\alpha=\dfrac{\sin\alpha}{\cos\alpha}=\dfrac{-3}{4}$
$\cot\alpha=\dfrac{1}{\tan\alpha}=\dfrac{-4}{3}$