Given that

$f\left(x\right)=\left\{\begin{array}{l}{x}^2+ax+b,0\le x<2\\ 3x+2,2\le x\le 4\\ 2ax+5b,4<x\le 8\end{array}\right.$

Since $f\left(x\right)$ is continuous on $\left[0,8\right]$ i.e. $f\left(x\right)$ is continuous at $x=2$ and $x=4$

$L.H.L=f\left(x\right)=R.H.L$

$\underset{x\to 2}{\lim }f\left(x\right)=f\left(2\right)$

$\underset{x\to 2}{\lim }\left(x^2+ax+b\right)=3x+2=8$

$4+2a+b=6+2$

$or 2a+b=4 \left(i\right)....$

$f\left(4\right)=\underset{x\to 4}{\lim f\left(x\right)}$

$3x+4=3\left(4\right)+2=\underset{x\to 4}{\lim }\left(2ax+5b\right)$

$14=8a+5b \left(ii\right).....$

Solving $\left(i\right) \& \left(ii\right)$

$a=3 \& b=-2$