1. Given function

$f(x)=\ln(4x^2+11x+6)+\sin^{-1}(4x+3)+\cos^{-1}\left(\frac{10x+6}{3}\right)$

We need the domain of this function.

For the domain, all three terms must be defined simultaneously.

2. Condition from the logarithm

For $\ln(4x^2+11x+6)$ to be defined,

$4x^2+11x+6>0$

Factorize:

$4x^2+11x+6=(4x+3)(x+2)$

So we need

$(4x+3)(x+2)>0$

Critical points are $x=-2$ and $x=-\frac34$.

Thus,

$x\in(-\infty,-2)\cup\left(-\frac34,\infty\right)$

3. Condition from $\sin^{-1}(4x+3)$

For $\sin^{-1}(4x+3)$ to be defined,

$-1\le 4x+3\le 1$

Subtract $3$:

$-4\le 4x\le -2$

Divide by $4$:

$-1\le x\le -\frac12$

4. Condition from $\cos^{-1}\left(\frac{10x+6}{3}\right)$

For $\cos^{-1}(y)$ to be defined,

$-1\le y\le 1$

So,

$-1\le \frac{10x+6}{3}\le 1$

Multiply by $3$:

$-3\le 10x+6\le 3$

Subtract $6$:

$-9\le 10x\le -3$

Divide by $10$:

$-\frac9{10}\le x\le -\frac3{10}$

5. Intersect all conditions

We need

$x\in\left[ -1,-\frac12\right] \cap \left[-\frac9{10},-\frac3{10}\right] \cap \left[(-\infty,-2)\cup\left(-\frac34,\infty\right)\right]$

First intersect the inverse trigonometric conditions:

$\left[ -1,-\frac12\right] \cap \left[-\frac9{10},-\frac3{10}\right]=\left[-\frac9{10},-\frac12\right]$

Now intersect with the logarithm condition:

$\left[-\frac9{10},-\frac12\right] \cap \left[(-\infty,-2)\cup\left(-\frac34,\infty\right)\right]$

Since $\left[-\frac9{10},-\frac12\right]$ does not meet $(-\infty,-2)$, only the second part matters:

$\left[-\frac9{10},-\frac12\right] \cap \left(-\frac34,\infty\right)=\left(-\frac34,-\frac12\right]$

So the domain is

$(\alpha,\beta]=\left(-\frac34,-\frac12\right]$

Hence,

$\alpha=-\frac34,\qquad \beta=-\frac12$

6. Compute $36|\alpha+\beta|$

$\alpha+\beta=-\frac34-\frac12=-\frac34-\frac24=-\frac54$

Therefore,

$|\alpha+\beta|=\frac54$

So,

$36|\alpha+\beta|=36\cdot \frac54=9\cdot 5=45$

7. Option check

The value is

$\boxed{45}$

So the correct option is C.