$f\left(x\right)=$$\left\{\begin{array}{cc}\frac{{(x+b{x}^2)}^{\frac{1}{2}}-x^{\frac{1}{2}}}{b{x}^{\frac{1}{2}}}& x>0\\ c& x=0\\ \frac{\sin (a+1)x+\sin x}{x}& x<0\end{array}\right\}$ is continuous at  $x=0$ then

$\underset{x\to 0^{+}}{\lim }\frac{\sqrt{x+b{x}^2}-\sqrt{x}}{b\sqrt{x}}=\underset{x\to 0^{+}}{\lim }\frac{\sqrt{x(1+bx)}-\sqrt{x}}{b\sqrt{x}}$ 

$=\underset{x\to 0^{+}}{\lim }\sqrt{x}\frac{(\sqrt{1+bx}-1)}{b\sqrt{x}}=\frac{0}{b}=0$

 $\therefore f\left(x\right)$ is continuous at $x=0$implies that  $\begin{array}{c}c=0\end{array}$and $\begin{array}{c}b\ne 0\end{array}$

$\underset{x\to 0^{-}}{\lim }\frac{\sin (a+1)x+\sin x}{x}=\underset{x\to 0^{-}}{\lim }(a+1)\cos (a+1)x+\cos x$, by L-Hospital rule.

$=a+1+1=0$

$\Rightarrow \begin{array}{c}a=-2\end{array}$