\(\pi< \alpha< \dfrac{3\pi}{2}\Rightarrow sin\alpha< 0\)\(\Rightarrow sin\alpha=-\sqrt{1-cos^2\alpha}=\dfrac{-\sqrt{15}}{4}\)
\(sin\left(\alpha-\dfrac{2\pi}{3}\right)=sin\alpha.cos\dfrac{2\pi}{3}-cos\alpha.sin\dfrac{2\pi}{3}\)
\(=\dfrac{-\sqrt{15}}{4}.\dfrac{-1}{2}-\dfrac{-1}{4}.\dfrac{\sqrt{3}}{2}=\dfrac{\sqrt{15}+\sqrt{3}}{8}\)