5. For each value of the parameter $a$, solve the equation $\log _{2} \frac{3 \sqrt{3}+\cos a(\sin x+4)}{3 \sin a \cos x}=|3 \sin a \cos x|-|\cos a(\sin x+4)+3 \sqrt{3}|$.
Solution: Domain of definition (ODZ): $\frac{3 \sqrt{3}+\cos a(\sin x+4)}{3 \sin a \cos x}>0$. On the ODZ, we have
$\log _{2} \frac{3 \sqrt{3}+\cos a(\sin x+4)}{3 \sin a \cos x}=\log _{2}|\cos a(\sin x+4)+3 \sqrt{3}|-\log _{2}|3 \sin a \cos x|$. Then the original equation on the ODZ is equivalent to the following $\log _{2}|\cos a(\sin x+4)+3 \sqrt{3}|+|\cos a(\sin x+4)+3 \sqrt{3}|=\log _{2}|3 \sin a \cos x|+|3 \sin a \cos x|$. Consider the function $f(t)=\log _{2} t+t$. The function $f(t)$ is increasing on the interval $(0 ;+\infty)$. Let $u=|\cos a(\sin x+4)+3 \sqrt{3}|, \quad v=|3 \sin a \cos x|$. Then we have the equation $f(u)=f(v)$, and due to the strict monotonicity of the function $f$, we arrive at the equation $u=v$, i.e., $|\cos a(\sin x+4)+3 \sqrt{3}|=|3 \sin a \cos x|$. The last equation on the ODZ is equivalent to the following $\cos a(\sin x+4)+3 \sqrt{3}=3 \sin a \cos x . \quad$ We have $\quad \cos a \sin x-3 \sin a \cos x=-3 \sqrt{3}-4 \cos a \quad \Leftrightarrow$ $\frac{\cos a}{\sqrt{8 \sin ^{2} a+1}} \sin x-\frac{3 \sin a}{\sqrt{8 \sin ^{2} a+1}} \cos x=-\frac{3 \sqrt{3}+4 \cos a}{\sqrt{8 \sin ^{2} a+1}} \Leftrightarrow \quad \sin (x-\varphi)=-\frac{3 \sqrt{3}+4 \cos a}{\sqrt{8 \sin ^{2} a+1}}, \quad$ where $\cos \varphi=\frac{\cos a}{\sqrt{8 \sin ^{2} a+1}}, \sin \varphi=\frac{3 \sin a}{\sqrt{8 \sin ^{2} a+1}}$ (auxiliary angle formula). For the equation to have solutions, the condition $-1 \leq-\frac{3 \sqrt{3}+4 \cos a}{\sqrt{8 \sin ^{2} a+1}} \leq 1$ must be satisfied. Since $3 \sqrt{3}+4 \cos a \geq 3 \sqrt{3}-4>0,-\frac{3 \sqrt{3}+4 \cos a}{\sqrt{8 \sin ^{2} a+1}}<0 \leq 1$, the right inequality is satisfied for any values of $a$. The left inequality is equivalent to the following $\sqrt{8 \sin ^{2} a+1} \geq 3 \sqrt{3}+4 \cos a \Leftrightarrow$ $8 \sin ^{2} a+1 \geq 27+24 \sqrt{3} \cos a+16 \cos ^{2} a \Leftrightarrow \quad 24 \cos ^{2} a+24 \sqrt{3} \cos a+18 \leq 0 \Leftrightarrow(2 \cos a+\sqrt{3})^{2} \leq 0 \Leftrightarrow$ $\cos a=-\frac{\sqrt{3}}{2} \Leftrightarrow a= \pm \frac{5 \pi}{6}+2 \pi n, n \in Z$.

1) Substitute the value $a=\frac{5 \pi}{6}+2 \pi n, n \in Z$ into the equation $\cos a \sin x-3 \sin a \cos x=-3 \sqrt{3}-4 \cos a$. We get $\quad-\frac{\sqrt{3}}{2} \sin x-\frac{3}{2} \cos x=-3 \sqrt{3}+2 \sqrt{3} \Leftrightarrow \sin \left(x+\frac{\pi}{3}\right)=1 \Leftrightarrow x=\frac{\pi}{6}+2 \pi k, k \in Z$. The only condition for checking the domain of definition is $\sin a \cos x \neq 0$, which is satisfied.
2) Substitute the value $a=-\frac{5 \pi}{6}+2 \pi n, n \in Z$ into the equation $\cos a \sin x-3 \sin a \cos x=-3 \sqrt{3}-4 \cos a$. We get $\quad-\frac{\sqrt{3}}{2} \sin x+\frac{3}{2} \cos x=-3 \sqrt{3}+2 \sqrt{3} \Leftrightarrow \sin \left(x-\frac{\pi}{3}\right)=1 \Leftrightarrow x=\frac{5 \pi}{6}+2 \pi k, k \in Z$. The only condition for checking the domain of definition is $\sin a \cos x \neq 0$, which is satisfied.

Answer: for $a=\frac{5 \pi}{6}+2 \pi n, n \in Z$ solutions $\quad x=\frac{\pi}{6}+2 \pi k, k \in Z$; for $a=-\frac{5 \pi}{6}+2 \pi n, n \in Z$ solutions $x=\frac{5 \pi}{6}+2 \pi k, k \in Z$; for $a \neq \pm \frac{5 \pi}{6}+2 \pi n, n \in Z$, there are no solutions.