Number of solution of $\tan \left(2x\right)=\tan \left(6x\right)in(0,3\pi )$ is

$\tan 2x=\tan 6x\Rightarrow \sin 4x=0$

$4x=\pi ,2\pi ,3\pi ,\cdots ,11\pi$

$x=\frac{\pi }{4},\frac{2\pi }{4},\frac{3\pi }{4},\cdots ,\frac{11\pi }{4}$

But $\frac{\pi }{4},\frac{3\pi }{4},\frac{5\pi }{4},\frac{7\pi }{4},\frac{9\pi }{4},\frac{11\pi }{4}$ are rejected

So number of solutions $=5$