The angle of intersection of the curves $y=4-x^2$ and $y=x^2$ is

$\tan ^{-1}\left(\frac{4 \sqrt{2}}{7}\right)$

Solving $y=4-x^2$ and $y=x^2$, the point of intersection is $(\sqrt{2}, 2)$

$m_1=\text { slope of tangent }=2 \sqrt{2}$

$m_2=\text { slope of tangent }=-2 \sqrt{2}$

∴  $\tan \theta=\left|\frac{4 \sqrt{2}}{1-8}\right|=\frac{4 \sqrt{2}}{7}$.