If the value of

$\lim _{x \rightarrow 0}(2-\cos x \sqrt{\cos } x)^{\frac{x+2}{x^{2}}}$

form: $1^{\infty}$

$=e^{\lim _{x \rightarrow 0}\left(\frac{1-\cos x \sqrt{\cos 2 x}}{x^{2}}\right) \times(x+2)}$

Now $\lim _{x \rightarrow 0} \frac{1-\cos x \sqrt{\cos 2 x}}{x^{2}}$

$=$ $\operatorname{limt}_{x \rightarrow 0} \frac{\sin x \sqrt{\cos 2 x}-\cos x \times \frac{1}{2 \sqrt{\cos 2 x}} \times(-2 \sin 2 x)}{2 x}$

(by L' Hospital Rule)

$\lim _{x \rightarrow 0} \frac{\sin x \cos 2 x+\sin 2 x \cdot \cos x}{2 x}$

$=\frac{1}{2}+1=\frac{3}{2}$

So, $e^{\lim _{x \rightarrow 0}\left(\frac{1-\cos x \sqrt{\cos 2 x}}{x^{2}}\right)(x+2)}$

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

$\Rightarrow a=3$