$$ \mathop {\lim }\limits_{x \to 0} \frac{1-\cos \left(x^2+\pi(x+2)\right)}{x^2}= $$

The value of ' $a$ ' for which the function

$f(x)=\left\{\begin{array}{cl}\frac{1-\cos 4 x}{x^2}, & x<0 \\ \frac{a}{\sqrt{x}}, & x=0 \text { is continuous at } x=0, \text { is } \\ \frac{\sqrt{16+\sqrt{x}}-4}{\sqrt{16+}} & \end{array}\right.$

$$ \begin{aligned} & \text { If } f(x)=\tan ^{-1}\left(\frac{1}{\sin ^2 x+\sin x+1}\right) \\ & \quad+\tan ^{-1}\left(\frac{1}{\sin ^2 x+3 \sin x+3}\right)+\tan ^{-1} \end{aligned} $$

$\left(\frac{1}{\sin ^2 x+5 \sin x+7}\right)+\ldots+$ upto 10 terms, then $f^{\prime}(0)=$

If $\alpha$ is such a minimum value for which the inverse of $f(x)=x^2+3 x-3$ exists in $[\alpha, \infty)$ and $g$ is the inverse of the $f$, then at $x=\alpha+\frac{5}{2}, \frac{d g}{d x}$