If $f(x)=\left\{\begin{array}{cl}\frac{\left(e^{a x}-1\right) \log (1+x)}{\sin ^2 x}, & \text { if } x>0 \\ 2, & \text { if } x=0 \\ \frac{\cos 4 x-\cos b x}{\tan ^2 x}, & \text { if } x<0\end{array}\right.$ is continuous at $x=0$, then $\sqrt{b^2-a^2}=$

$$\mathop {\lim }\limits_{x \to 0} \frac{x^2 \sin ^2(3 x)+\sin ^4(6 x)}{(1-\cos 3 x)^2}= $$

If a real valued function

$$ f(x)=\left\{\begin{array}{cc} (1+\sin x)^{\cos x}, & -\pi / 2 < x < 0 \\ a, & x=0 \\ \frac{e^{2 / x}+e^{3 / x}}{a e^{2 / x}+b e^{3 / x}}, & 0 < x < \pi / 2 \end{array}\right. $$

is continuous at $x=0$, then $a b=$

$$ \mathop {\lim }\limits_{x \to 0} \frac{(\operatorname{cosec} x-\cot x)\left(e^x-e^{-x}\right)}{\sqrt{3}-\sqrt{2+\cos x}}= $$