The acute angle between the tangents drawn at the point of intersection (other than the origin) of the curves $x^2=4y$ and $y^2=4x$ is 

$x^2=4y\to \left(1\right)\&y^2=4x\to \left(2\right)$

Solve (1) & (2) $\Rightarrow (0,0)\&(4,4)$

Diff . Eq (1)

$2x=4\frac{dy}{dx}$

$m_1={\left(\frac{dy}{dx}\right)}_{(4,4)}=2$

Diff. Eq (2)

$2y\frac{dy}{dx}=4\left(1\right)$

$m_2={\left(\frac{dy}{dx}\right)}_{(4,4)}=\frac{1}{2}$

$\tan \theta =\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|=\frac{3}{4}$

$\Rightarrow \sin \theta =\frac{3}{5}\Rightarrow \theta ={\sin }^{-1}(3/5)$