Do $\dfrac{\pi}{2} < \alpha < \pi$
$\Rightarrow \begin{cases} \tan \alpha >0\\ \cos \alpha >0 \end{cases}$
$\Rightarrow \cos \alpha = \sqrt{1-\Big(\dfrac{3}{4}\Big)^2}$
$= \dfrac{\sqrt{7}}{4}$
$\to \tan \alpha = \dfrac{\dfrac{3}{4}}{\dfrac{\sqrt{7}}{4}}= \dfrac{3\sqrt{7}}{7}$
$\cos \Big(\alpha + \dfrac{\pi}{3} \Big)$
$= \cos \alpha \cos \dfrac{\pi}{3} - \sin \alpha \sin \dfrac{\pi}{3}$
$= \dfrac{\sqrt{7}}{4}. \dfrac{1}{2} - \dfrac{3}{4} \dfrac{\sqrt{3}}{2}$
$= \dfrac{\sqrt{7} - 3\sqrt{3}}{8}$
