$$ \mathop {\lim }\limits_{x \to 1}\left(\log _3 3 x\right)^{\log _x 8}=\ldots $$

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

is continuous at $x=0$, then $\mathrm{a}=$

If the function

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

is continuous in $[0, \pi]$ then $a-b=$

$$ \lim\limits_{x \rightarrow \infty}\left(\frac{x+8}{x+1}\right)^{x+5}=\ldots $$