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Trả lời:
$\dfrac{\pi}{2}<a<\pi⇒\begin{cases}\sin a>0\\\cos<0\end{cases}$
$\sin a=\dfrac{4}{5}⇒\cos a=-\sqrt{1-\sin^2a}=-\dfrac{3}{5}$
$\sin\bigg{(}a+\dfrac{\pi}{3}\bigg{)}=\sin a.\cos\dfrac{\pi}{3}+\cos a.\sin\dfrac{\pi}{3}=\dfrac{4}{5}.\dfrac{1}{2}-\dfrac{3}{5}.\dfrac{\sqrt{3}}{2}=\dfrac{4-3\sqrt{3}}{10}$.