Given $\frac{3}{4}$ is a root of $a x^{2}+b x-6=0$;
therefore, we have:

$a \times\left(\frac{3}{4}\right)^{2}+b \times \frac{3}{4}-6=0$
$\begin{array}{l}
\Rightarrow \frac{9 a}{16}+\frac{3 b}{4}=6 \\
\Rightarrow \frac{9 a+12 b}{16}=6 \\
\Rightarrow 9 a+12 b-96=0 \\
\Rightarrow 3 a+4 b=32
\end{array}$ …..(i)

Again, $(-2)$ is a root of $a x^{2}+b x-6=0$ so,

$\begin{array}{l}
a \times(-2)^{2}+b \times(-2)-6=0 \\
\Rightarrow 4 a-2 b=6 \\
\Rightarrow 2 a-b=3
\end{array}$ …..(ii)

On multiplying (ii) by 4 and adding the result with (i), we get:

$\begin{array}{l}
\Rightarrow 3 a+4 b+8 a-4 b=32+12 \\
\Rightarrow 11 a=44 \\
\Rightarrow a=4
\end{array}$

Putting the value of a in (ii), we get:

$\begin{array}{l}
2 \times 4-b=3 \\
\Rightarrow 8-b=3 \\
\Rightarrow b=5
\end{array}$

As a result, the required values of $a$ and $b$ are 4 and 5, respectively.