Given, $4 x-3 y=5$ and $2 x^2-3 y^2=12$
$\therefore \quad 2\left(\frac{5+3 y}{4}\right)^2-3 y^2=12$
$\begin{aligned} & \Rightarrow \quad \frac{\left(25+9 y^2+30 y\right)}{8}-3 y^2=12 \\ & \Rightarrow \quad 15 y^2-30 y+71=0 \\ & \Rightarrow \quad y=\frac{30 \pm \sqrt{900-4260}}{30} \\ & =1 \pm \frac{\sqrt{-3360}}{30} \\ & \end{aligned}$
Also,
$\begin{aligned}
& 2 x^2-3\left(\frac{4 x-5}{3}\right)^2=12 \\
& \Rightarrow \quad 10 x^2-40 x+61=0 \\
&
\end{aligned}$
$\Rightarrow \quad x=\frac{40 \pm \sqrt{1600-4 \times 10 \times 61}}{2 \times 10}$
$=\frac{40 \pm \sqrt{-840}}{20}=2 \pm \frac{\sqrt{-840}}{20}$
$\therefore \quad$ Points are $A\left(2+\frac{\sqrt{-840}}{20}, 1+\frac{\sqrt{-3360}}{30}\right)$ and
$B\left(2-\frac{\sqrt{-840}}{20}, 1-\frac{\sqrt{-3360}}{30}\right) \text {. }$
$\therefore \quad$ Mid point of $A B$ is $(2,1)$.