$\begin{array}{l}a)\quad \cos\left(2x + \dfrac{\pi}{3}\right) + \sin\left(2x + \dfrac{\pi}{3}\right) = \dfrac{\sqrt6}{2}\\ \Leftrightarrow \dfrac{\sqrt2}{2}\cdot\cos\left(2x + \dfrac{\pi}{3}\right) + \dfrac{\sqrt2}{2}\cdot\sin\left(2x + \dfrac{\pi}{3}\right) = \dfrac{\sqrt3}{2}\\ \Leftrightarrow \cos\left(2x + \dfrac{\pi}{3} - \dfrac{\pi}{4}\right) = \cos\dfrac{\pi}{6}\\ \Leftrightarrow \cos\left(2x - \dfrac{\pi}{12}\right) = \cos\dfrac{\pi}{6}\\ \Leftrightarrow \left[\begin{array}{l}2x - \dfrac{\pi}{12} = \dfrac{\pi}{6} +k2\pi\\2x - \dfrac{\pi}{12} = -\dfrac{\pi}{6} + k2\pi\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}2x = \dfrac{\pi}{4} +k2\pi\\2x =- \dfrac{\pi}{12} + k2\pi\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x = \dfrac{\pi}{8} +k\pi\\x =- \dfrac{\pi}{24} + k\pi\end{array}\right.\quad (k \in \Bbb Z)\\ b)\quad (2\sin x +1)(2\cos2x + \sin x +1) = 3-4\cos^2x\\ \Leftrightarrow (2\sin x + 1)[2(1 - 2\sin^2x) + \sin x +1] = 3 - 4(1 - \sin^2x)\\ \Leftrightarrow (2\sin x +1)(-4\sin^2x + \sin x +3) = 4\sin^2x - 1\\ \Leftrightarrow (2\sin x + 1)(-4\sin^2x + \sin x +3)= (2\sin x +1)(2\sin x -1)\\ \Leftrightarrow (2\sin x +1)(4\sin^2x+ \sin x -4) = 0\\ \Leftrightarrow \left[\begin{array}{l}2\sin x + 1 = 0\\4\sin^2x+ \sin x -4=0\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}\sin x = -\dfrac12\\\sin x =\dfrac{-1 -\sqrt{65}}{8}\quad (loại)\\\sin x = \dfrac{-1+\sqrt{65}}{8} \end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x = - \dfrac{\pi}{6} +k2\pi\\x = \dfrac{7\pi}{6} + k2\pi\\x = \arcsin\left(\dfrac{-1+\sqrt{65}}{8}\right) + k2\pi\\x = \pi - \arcsin\left(\dfrac{-1+\sqrt{65}}{8}\right) + k2\pi\end{array}\right.\quad (k \in \Bbb Z) \end{array}$
