$\text{ Let }g\left(x\right)=\left[\begin{array}{ll}a\sqrt{x+1}& \text{ if }0<x<3 \text{ if }g\left(x\right)\text{ is differentiable on }(0,5) then a=\\ bx+2& \text{ if }3\le x<5\end{array}\right.$

$g^{\text{'}}\left(3^{-}\right)=\underset{h\to 0}{Lt}\frac{g\left(3-h\right)-g\left(3\right)}{-h}=\underset{h\to 0}{Lt}\frac{a\sqrt{4-h}-\left(3b+2\right)}{-h}$

$\begin{array}{l}\text{ For existence of limit} numerator must be zero at h = 0\\ \therefore 2a-3b=2\\ g^{\text{'}}\left(3^{+}\right)=L_{h\to 0}t\frac{b(3+h)+2-(3b+2)}{h}=b\end{array}$

$\begin{array}{l}\text{ Substituting }3b+2=2a\\ \therefore g^{\text{'}}\left(3^{-}\right)=\underset{h\to 0}{lim}\frac{a\sqrt{4-h}-2a}{-h}=\frac{a}{4}\end{array}$

$\begin{array}{l}\therefore g^{\text{'}}\left(3^{-}\right)=g^{\text{'}}\left(3^{+}\right)\\ \frac{a}{4}=b\\ substitute this in the equation 2a-3b=2\\ 2a-3\frac{a}{4}=2\\ 8a-3a=8\\ 5a=8\\ a=\frac{8}{5}\\ \end{array}$