If the point $P(a^2,a)$ lies in the region corresponding to the acute angle between the lines $2y=x$ and $4y=x$ , then

But $a^2>0$, hence point $P(a^2,a)$ lies in first quadrant. We have

$a-\left(\frac{a^2}{4}\right)>0 \& a-\left(\frac{{\displaystyle a^2}}{{\displaystyle 2}}\right)<0$

($\therefore (1,0)$ and P lies on same side of x-2y=0 and (1,0) and P lies opposite sides of x-4y=0)

$\Rightarrow 0<a and a\in (-\infty ,0)\cup (2,\infty )$

$\Rightarrow a\in (2,4)$