If a function $f$ defined by

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

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

$$ \mathop {\lim }\limits_{x \to \infty } \frac{(\sqrt{2})-\sqrt{1+\cos x}}{\sqrt{15+\cos 2 x-4}}= $$

If a real valued function

$$ f(x)=\left\{\begin{array}{cl} \frac{x^2+(a+3) x+(a+1)}{x+3} & , \text { when } x \neq-3 \\ -\frac{5}{2} & , \text { when } x=-3 \end{array}\right. $$

is continuous at $x=-3$, then $\lim _{x \rightarrow a}\left(x^2+x+1\right)=$

$$ \mathop {\lim }\limits_{x \to 0} \frac{x \tan 2 x-2 x \tan x}{(1-\cos 3 x)(\operatorname{cosec} x-\cot x)^2}= $$