$$\lim _\limits{x \rightarrow 0} \frac{x \cot 4 x}{\sin ^2 x \cdot \cot ^2(2 x)} \text { is equal to }$$

Given $$\mathrm{f}(x)=\left\{\begin{array}{cc}\frac{1-\cos 4 x}{x^2} & , \text { if } x<0 \\ \mathrm{a} & , \text { if } x=0 \\ \frac{\sqrt{x}}{\sqrt{16-\sqrt{x}-4}}, & \text { if } x>0\end{array}\right.$$

If $$\mathrm{f}(x)$$ is continuous at $$x=0$$, then value of a is

The values of $$a$$ and $$b$$, so that the function

$$f(x)=\left\{\begin{array}{l} x+a \sqrt{2} \sin x, 0 \leq x \leq \frac{\pi}{4} \\ 2 x \cot x+b, \frac{\pi}{4} \leq x \leq \frac{\pi}{2} \\ a \cos 2 x-b \sin x, \frac{\pi}{2} < x \leq \pi \end{array}\right.$$

is continuous for $$0 \leq x \leq \pi$$, are respectively given by

$$\lim _\limits{x \rightarrow 0} \frac{\cos 7 x^{\circ}-\cos 2 x^{\circ}}{x^2}$$ is