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