Answer
$a=3,\ m=1,\ b=4.$
Work Step by Step
The two hypotheses of the Mean Value Theorem are:
(1) $y=f(x)$ is continuous over a closed interval $[a, b]$ and
(2) $f(x)$ is differentiable on the interval's interior $(a, b)$.
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In order for the function to satisfy (1), we must have
$\displaystyle \lim_{x\rightarrow 0^{+}}(-x^{2}+3x+a)=3\ \Rightarrow\ a=3$
and
$\displaystyle \lim_{x\rightarrow 1^{-}}(-x^{2}+3x+a)=\lim_{x\rightarrow 1^{+}}(mx+b) \ $
$\Rightarrow\ -1+3+3=m+b$
$\Rightarrow\ 5=m+b$
In order to satisfy (2), we also must have
$\displaystyle \left.\frac{d}{dx}(-x^{2}+3x+a)\right|_{x=1}=\left.\frac{d}{dx}(mx+b)\right|_{x=1}$
$\Rightarrow\ \left. (-2x+3)\right|_{x=1}=m$
$m=1,$
Back-substitute: $b=5-m=4.$
