$\frac{2 x-3}{4 x+5}>0$

$\begin{aligned} & \therefore \quad x \in\left(-\infty,-\frac{5}{4}\right) \cup\left(\frac{3}{2}, \infty\right) ........(i) \\ & -1 \leq \frac{3 x+4}{2-x} \leq 1 \\ & \frac{3 x+4}{2-x} \leq 1 \\ & \Rightarrow \quad \frac{3 x+4}{2-x}-1 \leq 0\end{aligned}$

$\begin{aligned} & \Rightarrow \quad \frac{3 x+4-2+x}{x-2} \geq 0 \\ & \Rightarrow \quad \frac{4 x+2}{x-2} \geq 0 \\ & \Rightarrow \quad x \in\left(-\infty,-\frac{1}{2}\right] \cup(2, \infty)........(ii)\end{aligned}$

$$ \begin{aligned} & \frac{3 x+4}{2-x} \geq-1 \\ \Rightarrow & \frac{3 x+4}{2-x}+1 \geq 0 \\ \Rightarrow & \frac{3 x+4+2-x}{2-x} \geq 0 \\ \Rightarrow & \frac{2 x+6}{x-2} \leq 0 \\ \therefore & x \in[-3,2)........(iii) \end{aligned} $$

Taking intersection of (i), (ii) and (iii)

$$ \begin{aligned} & x \in\left[-3,-\frac{5}{4}\right) \\ & \alpha=-3, \beta=-\frac{5}{4} \\ & \alpha^2+4 \beta=4 \end{aligned} $$