Let $f:[-1,3] \rightarrow \mathbb{R}$ be defined as

$$\left\{\begin{array}{lc} |x|+[x], & -1 \leqslant x<1 \\ x+|x|, & 1 \leqslant x<2 \\ x+[x], & 2 \leqslant x \leqslant 3 \end{array}\right.$$

where $[t]$ denotes the greatest integer function. Then $f$ is discontinuous at

$$\lim _\limits{x \rightarrow \frac{\pi}{2}} \frac{\left(1-\tan \left(\frac{x}{2}\right)\right)(1-\sin x)}{\left(1+\tan \left(\frac{x}{2}\right)\right)(\pi-2 x)^3}$$ is

$\lim _\limits{x \rightarrow 0} \frac{x \tan 2 x-2 x \tan x}{(1-\cos 2 x)^2}$ is

Let $f(x)=\left\{\begin{array}{cc}\frac{1-\cos 4 x}{x^2} & , x<0 \\ a & , x=0 \\ \frac{\sqrt{2}}{\sqrt{16+\sqrt{x-4}}} & , x>0\end{array}\right.$ If $\mathrm{f}(x)$ is continuous at $x=0$, then the value of $a$ is