1. We need to evaluate

$$L=\lim_{x\to 0} \csc x\left(\sqrt{2\cos^2 x+3\cos x}-\sqrt{\cos^2 x+\sin x+4}\right).$$

2. First check the form at $x=0$:

• $\cos 0=1$
• $\sin 0=0$

So,

$$\sqrt{2\cos^2 0+3\cos 0}=\sqrt{2+3}=\sqrt5,$$

and

$$\sqrt{\cos^2 0+\sin 0+4}=\sqrt{1+0+4}=\sqrt5.$$

Thus the bracket is of the form $0$, while $\csc x\to \infty$, so overall it is a $\frac{0}{0}$-type situation.

3. Rationalize the expression inside the bracket:

$$L=\lim_{x\to 0}\frac{\sqrt{2\cos^2 x+3\cos x}-\sqrt{\cos^2 x+\sin x+4}}{\sin x}.$$

Multiply numerator and denominator by the conjugate:

$$L=\lim_{x\to 0} \frac{(2\cos^2 x+3\cos x)-(\cos^2 x+\sin x+4)}{\sin x\left(\sqrt{2\cos^2 x+3\cos x}+\sqrt{\cos^2 x+\sin x+4}\right)}.$$

4. Simplify the numerator:

$$(2\cos^2 x+3\cos x)-(\cos^2 x+\sin x+4) =\cos^2 x+3\cos x-\sin x-4.$$

Hence,

$$L=\lim_{x\to 0} \frac{\cos^2 x+3\cos x-\sin x-4}{\sin x\left(\sqrt{2\cos^2 x+3\cos x}+\sqrt{\cos^2 x+\sin x+4}\right)}.$$

5. Factor the numerator using

$$\cos^2 x+3\cos x-4=(\cos x-1)(\cos x+4).$$

So,

$$\cos^2 x+3\cos x-\sin x-4=(\cos x-1)(\cos x+4)-\sin x.$$

Therefore,

$$L=\lim_{x\to 0} \frac{(\cos x-1)(\cos x+4)-\sin x}{\sin x\left(\sqrt{2\cos^2 x+3\cos x}+\sqrt{\cos^2 x+\sin x+4}\right)}.$$

Split it:

$$L=\lim_{x\to 0} \frac{(\cos x-1)(\cos x+4)}{\sin x\left(\sqrt{2\cos^2 x+3\cos x}+\sqrt{\cos^2 x+\sin x+4}\right)} - \lim_{x\to 0} \frac{1}{\sqrt{2\cos^2 x+3\cos x}+\sqrt{\cos^2 x+\sin x+4}}.$$

6. Evaluate the first limit: As $x\to 0$,

$$\cos x-1\sim -\frac{x^2}{2},\qquad \sin x\sim x,$$

so

$$\frac{\cos x-1}{\sin x}\to 0.$$

Also $\cos x+4\to 5$, and the square-root sum tends to $2\sqrt5$. Hence the first term tends to $0$.

7. Evaluate the second limit directly:

\lim_{x\to 0}\frac{1}{\sqrt{2\cos^2 x+3\cos x}+\sqrt{\cos^2 x+\sin x+4}} =\frac{1}{\sqrt5+\sqrt5}=rac{1}{2\sqrt5}.

Since it is subtracted,

$$L=0-\frac{1}{2\sqrt5}=-\frac{1}{2\sqrt5}.$$

8. Therefore the correct option is

$$\boxed{-\frac{1}{2\sqrt5}}.$$

This is option D.