Giúp em với ạ, em không hiểu lắm ạ

$\sin \alpha = \dfrac{2}{3}$, $\dfrac{\pi}{2} < \alpha < \pi$.
Ta có $\sin^2 \alpha + \cos^2 \alpha = 1$
$\Leftrightarrow \left(\dfrac{2}{3}\right)^2 + \cos^2 \alpha = 1$
$\Leftrightarrow \dfrac{4}{9} + \cos^2 \alpha = 1$
$\Leftrightarrow \cos^2 \alpha = 1 - \dfrac{4}{9}$
$\Leftrightarrow \cos^2 \alpha = \dfrac{5}{9}$
Vì $\dfrac{\pi}{2} < \alpha < \pi$ nên $\cos \alpha < 0$.
$\Rightarrow \cos \alpha = -\dfrac{\sqrt{5}}{3}$.
a) $\cos \alpha < 0$ đúng
b) $\tan \alpha = \dfrac{\sin \alpha}{\cos \alpha} = \dfrac{\dfrac{2}{3}}{-\dfrac{\sqrt{5}}{3}} = -\dfrac{2}{\sqrt{5}} = -\dfrac{2\sqrt{5}}{5}$.
Vậy b) đúng.
c) $\cos\left(\dfrac{\pi}{3} + \alpha\right) = \cos\dfrac{\pi}{3}\cos\alpha - \sin\dfrac{\pi}{3}\sin\alpha$
$= \dfrac{1}{2} \cdot \left(-\dfrac{\sqrt{5}}{3}\right) - \dfrac{\sqrt{3}}{2} \cdot \dfrac{2}{3}$
$= -\dfrac{\sqrt{5}}{6} - \dfrac{2\sqrt{3}}{6}$
$= \dfrac{-\sqrt{5} - 2\sqrt{3}}{6}$.
Vậy c) đúng
d) 
$\sin 2\alpha = 2\sin\alpha\cos\alpha = 2 \cdot \dfrac{2}{3} \cdot \left(-\dfrac{\sqrt{5}}{3}\right) = -\dfrac{4\sqrt{5}}{9}$.
$\cos 2\alpha = \cos^2 \alpha - \sin^2 \alpha = \left(-\dfrac{\sqrt{5}}{3}\right)^2 - \left(\dfrac{2}{3}\right)^2 = \dfrac{5}{9} - \dfrac{4}{9} = \dfrac{1}{9}$.
$\sin\left(\dfrac{\pi}{4} - 2\alpha\right) = \sin\dfrac{\pi}{4}\cos 2\alpha - \cos\dfrac{\pi}{4}\sin 2\alpha$
$= \dfrac{\sqrt{2}}{2} \cdot \dfrac{1}{9} - \dfrac{\sqrt{2}}{2} \cdot \left(-\dfrac{4\sqrt{5}}{9}\right)$
$= \dfrac{\sqrt{2}}{18} + \dfrac{4\sqrt{10}}{18}$
$= \dfrac{\sqrt{2} + 4\sqrt{10}}{18}$.
Vậy d) là sai