Given that  $f\left(x\right)=\left\{\begin{array}{l}\sin x\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{ if }x\le 0\\ x^2+a\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{ if }0<x<1\\ bx+3\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{ if }1\le x\le 3\\ -3\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{ if }x>3\end{array}\right.$

Since f(x) is continuous at $x=0$ 

$\underset{x\to 0}{Lt}f\left(x\right)$ exists 

$\Rightarrow \underset{x\to 0+}{lt}f\left(x\right)=\underset{x\to 0-}{lt}f\left(x\right)$

$\Rightarrow \underset{x\to 0+}{lt}{x}^2+a=\underset{x\to 0-}{lt}\sin x$

$\Rightarrow 0+a=\sin 0=0\Rightarrow a=0$

$f\left(x\right)$ is continuous at $x=3$

$\Rightarrow \underset{x\to 3}{Lt}f\left(x\right)\text{ exists and }\underset{x\to 3^{+}}{lt}f\left(x\right)=\underset{x\to 3^{-}}{lt}f\left(x\right)$

$\Rightarrow \underset{x\to 3^{+}}{lt}-3=\underset{x\to 3^{-}}{lt}bx+3$

$\Rightarrow -3=b\left(3\right)+3\Rightarrow 3b=-6\Rightarrow b=-2$