Answer: 28.

Solution. Let $t=\frac{\pi x}{12}$, then the inequality takes the form

$$
\left(1-\operatorname{ctg}^{2} t\right)\left(1-3 \operatorname{ctg}^{2} t\right)(1-\operatorname{tg} 2 t \cdot \operatorname{ctg} 3 t) \leqslant 16
$$

Since

$$
\begin{gathered}
1-\operatorname{ctg}^{2} t=-\frac{\cos 2 t}{\sin ^{2} t}, \quad 1-3 \operatorname{ctg}^{2} t=\frac{\sin ^{2} t-3 \cos ^{2} t}{\sin ^{2} t}=\frac{4 \sin ^{2} t-3}{\sin ^{2} t}=-\frac{\sin 3 t}{\sin ^{3} t} \\
1-\operatorname{tg} 2 t \cdot \operatorname{ctg} 3 t=1-\frac{\sin 2 t \cdot \cos 3 t}{\cos 2 t \cdot \sin 3 t}=\frac{\sin t}{\cos 2 t \cdot \sin 3 t}
\end{gathered}
$$

the inequality reduces to $\frac{1}{\sin ^{4} t} \leqslant 16$, that is, $|\sin t| \geqslant \frac{1}{2}$. Therefore, $t \in\left[\frac{\pi}{6}+\pi n ; \frac{5 \pi}{6}+\pi n\right]$, $n \in \mathbb{Z}$. It is also necessary to take into account the conditions $\sin t \neq 0, \cos 2 t \neq 0, \sin 3 t \neq 0$. Thus, the points $\frac{\pi}{4}+\frac{\pi k}{2}$ and $\pm \frac{\pi}{3}+\pi k, k \in \mathbb{Z}$ need to be excluded from the obtained solution. Therefore, $x \in$ $[2+12 n ; 10+12 n], x \neq 3+6 n, x \neq \pm 4+12 n$. In the interval $[-3 ; 13]$, the numbers $-2,2,5,6,7$, 10 fall. Their sum is 28.
6.1. Find the sum of all integers \( x \in [-3 ; 13] \) that satisfy the inequality

$$
\left(1-\operatorname{ctg}^{2} \frac{\pi x}{12}\right)\left(1-3 \operatorname{ctg}^{2} \frac{\pi x}{12}\right)\left(1-\operatorname{tg} \frac{\pi x}{6} \cdot \operatorname{ctg} \frac{\pi x}{4}\right) \leqslant 16
$$