The function $f\left(x\right)$ $={\left(\sin 2x\right)}^{{\tan }^{2}2x}$  is not defined at $x=\pi /4$ , then the value of $f(\pi /4)$  so that $f$  is continuous at $x=\pi /4$  is

$f$  is continuous at $x=\pi /4$  if $\underset{x\to \pi /4}{\lim }$ $f\left(x\right)=f(\pi /4)$ 

Now $L=\underset{x\to \pi /4}{\lim }{\left(\sin 2x\right)}^{{\tan }^{2}2x}$

$\Rightarrow \log L=\underset{x\to \pi /4}{\lim }{\tan }^{2}2x\log \sin 2x$

                $=\underset{x\to \pi /4}{\lim }\frac{\log \sin 2x}{{\cot }^{2}2x}(\frac{0}{0} form u\sin g L\text{'}Hospital rule)$

                $=\underset{x\to \pi /4}{\lim }\frac{2\cot 2x}{-2\cot 2x\cos e{c}^{2}2x.2}$

                $=–\frac{1}{2}$

$\therefore$    $L=e^{-\frac{1^{ }}{2}}\Rightarrow f\left(\frac{\pi }{4}\right)=e^{-1/2}=\frac{1}{\sqrt{e}}$ .