Given $f(x)=\left\{\begin{array}{cc}\frac{x-[x]}{x-2} & x>2 \\ b & x=2 \\ \frac{\left|x^2-x-2\right|}{a\left(2+x-x^2\right)} & -1 < x \leq 2 \\ 2 a-b & x \leq-1\end{array}\right.$
Since $f(x)$ is continuous on $\mathrm{R}$
So, $\lim _{x \rightarrow 2^{+}} f(x)=f(2) \Rightarrow \lim _{x \rightarrow 2^{+}} \frac{x-[x]}{x-2}=b \Rightarrow b=1$
$\begin{aligned}
& \text { and } \lim _{x \rightarrow 2^{-}} f(x)=f(2) \Rightarrow \lim _{x \rightarrow 2^{-}} \frac{|(x-2)(x+1)|}{a(2-x)(x+1)}=1 \\
& \Rightarrow \lim _{x \rightarrow 2^{-}} \frac{-(x-2)(x+1)}{-a(x-2)) x+1)}=1 \Rightarrow a=1
\end{aligned}$
Now, $\lim _{x \rightarrow 0} \frac{\sin ^2 a x+x \tan b x}{x^2}=\lim _{x \rightarrow 0} \frac{\sin ^2 x+x \tan x}{x^2}$ $\Rightarrow \lim _{x \rightarrow 0}\left(\frac{\sin ^2 x}{x^2}+\frac{\tan x}{x}\right)=1+1=2$