$\frac{\pi}{2}<\alpha <\pi$
$\Rightarrow sin\alpha > 0$
$\Rightarrow sin\alpha= \sqrt{1-cos^2\alpha}= \frac{4}{5}$
$cos(\alpha-\frac{\pi}{3})= cos\alpha.cos\frac{\pi}{3}+sin\alpha.sin\frac{\pi}{3}$
$= \frac{-3}{5}.\frac{1}{2}+\frac{4}{5}.\frac{\sqrt{3}}{2}= \frac{4\sqrt{3}-3}{10}$