Given ${\sin }^{2}2\theta +{\cos }^{4}2\theta =\frac{3}{4}$

Using ${\sin }^{2}x+{\cos }^{2}x=1,$ we get

$\Rightarrow 1-{\cos }^{2}2\theta +{\cos }^{4}2\theta =\frac{3}{4}$

Let ${\cos }^{2}2\theta =t$

$\Rightarrow t^2-t+\frac{1}{4}=0$

$\Rightarrow {\left(t-\frac{1}{2}\right)}^2=0$

$\Rightarrow t=\frac{1}{2}$

$\Rightarrow {\cos }^{2}2\theta =\frac{1}{2}$

$\Rightarrow 2{\cos }^{2}2\theta -1=0$

Using $\cos 2A=2{\cos }^{2}A-1,$ we get

$\Rightarrow \cos 4\theta =0$

If $\cos x=0,$ then $x=\left(2n+1\right)\frac{\mathrm{\pi }}{2}, n\in Z.$

$\Rightarrow 4\theta =\left(2n+1\right)\frac{\pi }{2}$

$\Rightarrow \theta =\left(2n+1\right)\frac{\pi }{8}$

$\Rightarrow \theta =\frac{\pi }{8}, \frac{3\pi }{8} \in \left(0, \frac{\pi }{2}\right)$

Hence, the sum of the values of $\theta$ is $\frac{\mathrm{\pi }}{8}+\frac{3\mathrm{\pi }}{8}=\frac{\pi }{2}.$