Given,

$f\left(x\right)={\log }_e\left(\frac{2x+3}{4x^2+x-3}\right)+{\cos }^{-1}\left(\frac{2x-1}{x+2}\right)$

Now, using the definition of $\log$ function and ${\cos }^{-1}$ function we get,

$\frac{2x+3}{4x^2+x-3}>0$ and $-1\le \frac{2x-1}{x+2}\le 1$

$\Rightarrow \frac{2x+3}{(4x-3)(x+1)}>0, \frac{3x+1}{x+2}\ge 0 \& \frac{x-3}{x+2}\le 0$

 

Now, solving $\frac{2x+3}{(4x-3)(x+1)}>0$ we get,

$\Rightarrow x\in \left(\frac{-3}{2},-1\right)\cup \left(\frac{3}{4},\infty \right) ....\left(i\right)$

Similarly for $\frac{3x+1}{x+2}\ge 0$ we get,

$x\in (-\infty ,-2)\cup \left[\frac{-1}{3},\infty \right) ...\left(ii\right)$

And for $\frac{x-3}{x+2}\le 0$ we get,

$x\in (-2,3] ... \left(iii\right)$

Now, taking intersection of  $\left(i\right)\cap \left(ii\right)\cap \left(iii\right)$ we get,

$(\frac{3}{4},3]$

So, on comparing with $(\alpha ,\beta ]$ we get, $\alpha =\frac{3}{4} \& \beta =3$

Hence, $5\beta -4\alpha =15-3=12$