If the value of $$\lim_{x \to 0}\left(2 - \cos x\sqrt{\cos 2x}\right)^{\left(\frac{x+2}{x^2}\right)}$$ is equal to $$e^a$$, then $$a$$ is equal to ___.

We need to evaluate $$L = \lim_{x \to 0}\left(2 - \cos x\sqrt{\cos 2x}\right)^{(x+2)/x^2}$$. As $$x \to 0$$, the base approaches $$2 - 1 \cdot 1 = 1$$ and the exponent approaches $$\infty$$, giving the indeterminate form $$1^\infty$$.

We write $$L = e^A$$ where $$A = \lim_{x \to 0} \frac{x+2}{x^2}\left(2 - \cos x\sqrt{\cos 2x} - 1\right) = \lim_{x \to 0} \frac{x+2}{x^2}\left(1 - \cos x\sqrt{\cos 2x}\right)$$.

Using Taylor series: $$\cos x = 1 - \frac{x^2}{2} + O(x^4)$$ and $$\cos 2x = 1 - 2x^2 + O(x^4)$$. For the square root: $$\sqrt{\cos 2x} = (1 - 2x^2)^{1/2} \approx 1 - x^2 + O(x^4)$$. Multiplying:

$$\cos x \cdot \sqrt{\cos 2x} \approx \left(1 - \frac{x^2}{2}\right)\left(1 - x^2\right) = 1 - x^2 - \frac{x^2}{2} + \frac{x^4}{2} \approx 1 - \frac{3x^2}{2}$$

Therefore $$1 - \cos x\sqrt{\cos 2x} \approx \frac{3x^2}{2}$$ for small $$x$$. Substituting into the exponent:

$$A = \lim_{x \to 0} \frac{x+2}{x^2} \cdot \frac{3x^2}{2} = \lim_{x \to 0} \frac{3(x+2)}{2} = \frac{3 \cdot 2}{2} = 3$$

Therefore $$L = e^3$$, so $$a = 3$$.