We have,

$\underset{x\to 0}{\lim }{\left(2-\cos x\sqrt{\cos 2x}\right)}^{\left(\frac{x+2}{x^2}\right)}\to 1^{\infty }$

Hence,
$=e^{\underset{x\to 0}{\lim }\left(\frac{1-\cos x\sqrt{\cos 2\mathrm{x}}}{x^2}\right)\times \left(x+2\right)}$
Now,

$\underset{x\to 0}{\lim }\left(\frac{1-\cos x\sqrt{\cos 2\mathrm{x}}}{x^2}\right)\to \left(\frac{0}{0}\right)$

Using L'Hospital Rule

$\underset{x\to 0}{\lim }\left\{\frac{\sin x\sqrt{\cos 2x}-\cos x\times {\displaystyle \frac{1}{2\sqrt{\cos 2x}}}\times \left(-2\sin 2x\right)}{2x}\right\}$
 

$=\underset{x\to 0}{\lim }\left\{\frac{\sin x\cos 2x+\sin 2x·\cos x}{2x\sqrt{\cos 2x}}\right\}$

$=\underset{x\to 0}{\lim }\left\{\frac{\sin 3x}{2x\sqrt{\cos 2x}}\right\}$

$=\frac{3}{2}\underset{x\to 0}{\lim }\left\{\frac{\sin 3x}{3x}\right\}\times \underset{x\to 0}{\lim }\left\{\frac{1}{\sqrt{\cos 2x}}\right\}$

$=\frac{3}{2}\underset{x\to 0}{\lim }\left\{\frac{\sin 3x}{3x}\right\}\times 1$

$=\frac{3}{2}$

So,

$e^{\underset{x\to 0}{\lim }\left(\frac{1-\cos x\sqrt{\cos 2x}}{x^2}\right)\left(x+2\right)}$

$=e^{\frac{3}{2}\times 2}=e^3$

$\Rightarrow \boxed{a=3}$