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

Let f be twice differentiable function such that $\mathrm{f}^{\prime \prime}(x)=-\mathrm{f}(x), \mathrm{f}^{\prime}(x)=\mathrm{g}(x)$ and $\mathrm{h}(x)=(\mathrm{f}(x))^2+(\mathrm{g}(x))^2$. If $\mathrm{h}(5)=1$, then the value of $h(10)$ is

$$\lim _\limits{x \rightarrow 2}\left(\frac{5^x+5^{3-x}-30}{5^{3-x}-5^{\frac{x}{2}}}\right)=$$