$\dfrac{\pi}{2}<\alpha < \pi$
$\Rightarrow cos\alpha < 0$
$\Rightarrow cos\alpha = -\sqrt{1-sin^2\alpha}= -\dfrac{4}{5}$
$tan\alpha=\dfrac{sin\alpha}{cos\alpha}=-\dfrac{3}{4}$
$sin(\alpha+\dfrac{3\pi}{2})= sin\alpha.cos\dfrac{3\pi}{2}+cos\alpha.sin\dfrac{3\pi}{2}=\dfrac{4}{5}$
$cos(\alpha+\dfrac{-\pi}{2})= cos(\dfrac{\pi}{2}-\alpha)= sin\alpha=\dfrac{3}{5}$
$tan(\alpha-\pi)= -tan(\pi-\alpha)=tan\alpha= \dfrac{-3}{4}$