1. We need the domain of $f(x)=\ln\left(\frac{2x-3}{5+4x}\right)+\sin^{-1}\left(\frac{4+3x}{2-x}\right).$

The domain is the intersection of the domains of both parts.

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2. Domain from the logarithmic term $\ln\left(\frac{2x-3}{5+4x}\right)$ requires $\frac{2x-3}{5+4x}>0,$ and also $5+4x\neq 0$.

Critical points are: $2x-3=0 \Rightarrow x=\frac32,$ $5+4x=0 \Rightarrow x=-\frac54.$

Now check sign intervals:

• For $x< -\frac54$: numerator $2x-3<0$, denominator $5+4x<0$, so ratio $>0$.
• For $-\frac54 < x < \frac32$: numerator $<0$, denominator $>0$, so ratio $<0$.
• For $x>\frac32$: numerator $>0$, denominator $>0$, so ratio $>0$.

Hence log-domain is $(-\infty,-\tfrac54)\cup(\tfrac32,\infty).$

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3. Domain from the inverse sine term $\sin^{-1}\left(\frac{4+3x}{2-x}\right)$ requires $-1\le \frac{4+3x}{2-x}\le 1,$ with $x\ne 2$.

So solve both inequalities.

(i) Solve

$\frac{4+3x}{2-x}\le 1$ $\frac{4+3x-(2-x)}{2-x}\le 0$ $\frac{2+4x}{2-x}\le 0$ $\frac{1+2x}{2-x}\le 0.$ Critical points: $x=-\frac12,2$. This gives $x\in (-\infty,-\tfrac12]\cup(2,\infty)?$ Let us check carefully by sign chart:

• If $x< -\frac12$: numerator $1+2x<0$, denominator $2-x>0$, ratio $<0$ ✅
• If $-\frac12 < x <2$: numerator $>0$, denominator $>0$, ratio $>0$ ❌
• If $x>2$: numerator $>0$, denominator $<0$, ratio $<0$ ✅

Thus $\frac{1+2x}{2-x}\le 0 \Rightarrow x\in(-\infty,-\tfrac12]\cup(2,\infty).$

(ii) Solve

$\frac{4+3x}{2-x}\ge -1$ $\frac{4+3x+(2-x)}{2-x}\ge 0$ $\frac{6+2x}{2-x}\ge 0$ $\frac{x+3}{2-x}\ge 0.$ Critical points: $x=-3,2$.

Sign chart:

• If $x<-3$: numerator $<0$, denominator $>0$, ratio $<0$ ❌
• If $-3\le x<2$: numerator $\ge 0$, denominator $>0$, ratio $\ge 0$ ✅
• If $x>2$: numerator $>0$, denominator $<0$, ratio $<0$ ❌

Hence $\frac{x+3}{2-x}\ge 0 \Rightarrow x\in[-3,2).$

Intersecting (i) and (ii): $\left(( -\infty,-\tfrac12]\cup(2,\infty)\right)\cap[-3,2)=[-3,-\tfrac12].$

So the arcsin-domain is $[-3,-\tfrac12].$

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4. Intersect both domains:

Log-domain: $(-\infty,-\tfrac54)\cup(\tfrac32,\infty)$ Arcsin-domain: $[-3,-\tfrac12].$

Therefore common domain is $[-3,-\tfrac54).$

So $\alpha=-3,\qquad \beta=-\frac54.$

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5. Compute $\alpha^2+4\beta=(-3)^2+4\left(-\frac54\right)=9-5=4.$

Thus the correct option is $\boxed{4}.$