Solution

Firstly we can say that moduls of friction forces is the same(force of connecting and matereals are same):
$$F1 = F2 = F$$

Whats more in the end:
$$\omega1= \omega3= -\omega2$$
From second newton's law for rotation(for the first and second cylinders):

$$I\frac{\omega1 - \omega}{t} = -F\cdot R$$
$$I\frac{\omega2 - \omega}{t} = -2F\cdot R$$
Solving this system:

$$\omega2 = 2\omega1 - \omega$$

So we have:
$$\omega1=\omega3=\frac{\omega}{3}; \omega2=-\frac{\omega}{3}$$