Answer
\[a=\displaystyle\frac{1}{2}, b=\displaystyle\frac{1}{2}\]
Work Step by Step
\[f(x)=\left\{\begin{array}{ll}
\displaystyle\frac{x^2-4}{x-2}\;\;\;\;\;\;\;\;\;\;\;\; , x<2\\\\
ax^2-bx+3\;\;\; , 2\leq x<3\\\\
2x-a+b\;\;\;\;\;\;, x\geq 3
\end{array}\right.\]
Clearly $f(x)$ is continuous if $x\neq 2,3$. We will check the continuity at $x=2$ and $x=3$.
\[\lim_{x\rightarrow 2^{-}}f(x)=\lim_{x\rightarrow 2}\left(\frac{x^2-4}{x-2}\right)=\lim_{x\rightarrow 2}\left(\frac{(x-2)(x+2)}{x-2}\right)=\lim_{x\rightarrow 2}(x+2)=4\]
\[\lim_{x\rightarrow 2^{+}}f(x)=\lim_{x\rightarrow 2}(ax^2-bx+3)=4a-2b+3\]
If $f$ is continuous at $x=2$ then:
\[\lim_{x\rightarrow 2^{-}}f(x)=\lim_{x\rightarrow 2^{+}}f(x)\]
\[\Rightarrow 4=4a-2b+3\Rightarrow 4a-2b=1\;\;\;\;\;\ldots (1)\]
\[\lim_{x\rightarrow 3^{-}}f(x)=\lim_{x\rightarrow 3}\left(ax^2-bx+3\right)=9a-3b+3\]
\[\lim_{x\rightarrow 3^{+}}f(x)=\lim_{x\rightarrow 3}(2x-a+b)=6-a+b\]
If $f$ is continuous at $x=3$ then:
\[\lim_{x\rightarrow 3^{-}}f(x)=\lim_{x\rightarrow 3^{+}}f(x)\]
\[\Rightarrow 9a-3b+3=6-a+b\Rightarrow 10a-4b=3\;\;\;\;\;\ldots (2)\]
Multiply (1) by $-2$:
\[-8a+4b=-2\;\;\;\;\;\;\ldots (3)\]
Add (2) and (3):
\[2a=1\Rightarrow a=\frac{1}{2}\]
From equation (2):
\[4\left(\frac{1}{2}\right)-2b=1\Rightarrow 2b=1\Rightarrow b=\frac{1}{2}\]
Hence $a=\displaystyle\frac{1}{2}$ and $b=\displaystyle\frac{1}{2}$
