giải giúp mình câu này vs ạ, mik đag cần gấp

Đáp án+Giải thích các bước giải:

 Ta có: $\sin^{2}\alpha + \cos^{2}\alpha = 1⇒\sin^{2}\alpha = 1-\cos^{2}\alpha=1-\left(\dfrac{3}{5}\right)^{2}=\dfrac{16}{25}$

$⇒\sin\alpha = ±\sqrt{\dfrac{16}{25}}=±\dfrac{4}{5}$

Vì $0<\alpha<\dfrac{\pi}{2}$ nên $\sin\alpha > 0 ⇒ \sin\alpha = \dfrac{4}{5}$

+) $\sin\left(\alpha + \dfrac{\pi}{6}\right)$

$=\sin\alpha\cos\dfrac{\pi}{6} + \cos\alpha\sin\dfrac{\pi}{6}$

$=\dfrac{4}{5}.\dfrac{\sqrt{3}}{2} + \dfrac{3}{5}.\dfrac{1}{2}$

$=\dfrac{2\sqrt{3}}{5} + \dfrac{3}{10}$

$=\dfrac{4\sqrt{3}+3}{10}$

+) $\cos\left(\alpha - \dfrac{\pi}{3}\right)$

$=\cos\alpha\cos\dfrac{\pi}{3} + \sin\alpha\sin\dfrac{\pi}{3}$

$=\dfrac{3}{5}.\dfrac{1}{2} + \dfrac{4}{5}.\dfrac{\sqrt{3}}{2}$

$=\dfrac{4\sqrt{3}+3}{10}$

+) $\tan\left(\alpha + \dfrac{\pi}{4}\right)$

$=\dfrac{\sin\left(\alpha+\dfrac{\pi}{4}\right)}{\cos\left(\alpha+\dfrac{\pi}{4}\right)}$

$=\dfrac{\sin\alpha\cos\dfrac{\pi}{4}+\cos\alpha\sin\dfrac{\pi}{4}}{\cos\alpha\cos\dfrac{\pi}{4}-\sin\alpha\sin\dfrac{\pi}{4}}$

$=\dfrac{\dfrac{4}{5}.\dfrac{\sqrt{2}}{2}+\dfrac{3}{5}.\dfrac{\sqrt{2}}{2}}{\dfrac{3}{5}.\dfrac{\sqrt{2}}{2}-\dfrac{4}{5}.\dfrac{\sqrt{2}}{2}}$

$=\dfrac{\dfrac{2\sqrt{2}}{5}+\dfrac{3\sqrt{2}}{10}}{\dfrac{3\sqrt{2}}{10}-\dfrac{2\sqrt{2}}{5}}$

$=-7$