If $f\left(x\right)=\{\begin{array}{cc}\frac{\sin \left[x^2\right]\pi }{x^2-3x-18}+a{x}^3+b& 0\le x\le 1\\ 2\cos \pi x+{\tan }^{-1}x& 1<x\le 2\end{array}$ is differentiable function in  $\left[0,2\right]$, where $[.]$ denotes the greatest integer function, then

$$\begin{gathered} \sin \left[x^2\right]\pi =0 for 0\le x\le 1 \\ Hence, f\left(x\right)=\{\begin{array}{cc}a{x}^3+b,& 0\le x\le 1\\ 2\cos \pi x+{\tan }^{-1}x,& 1\le x\le 2\end{array} \end{gathered}$$

$f\left(x\right)$ is continuous and differentiable at  $x=1$

$$\begin{gathered} \therefore \underset{x\to 1^{-}}{\lim }f\left(x\right)=\underset{x\to 1^{+}}{\lim }f\left(x\right)=f\left(1\right) \\ a+b=-2+\frac{\pi }{4}=a+b \\ \Rightarrow a+b=-2+\frac{\pi }{4}\to \left(1\right) \end{gathered}$$

$$\begin{gathered} ALso, \underset{x\to 1^{-}}{\lim }\frac{f\left(x\right)-f\left(1\right)}{x-1}=\underset{x\to 1^{+}}{\lim }\frac{f\left(x\right)-f\left(1\right)}{x-1} \\ \Rightarrow \underset{x\to 1^{-}}{\lim }\frac{(a{x}^3+b)-(a+b)}{x-1}=\underset{x\to 1^{+}}{\lim }\frac{(2\cos \pi x+{\tan }^{-1}x)-(a+b)}{x-1} \\ \Rightarrow 3a=\underset{x\to 1^{+}}{\lim }-2\pi \sin \pi x+\frac{1}{1+x^2} \\ \Rightarrow 3a=\frac{1}{2} or a=\frac{1}{6} \\ \therefore b=\frac{\pi }{4}-\frac{13}{6} \end{gathered}$$