The no-slip condition requires the tangential velocities at the contact point $C$ to be equal for both circles. The velocity of point $C$ on circle A is $v_{CA} = \omega_1 R_1$. The center of circle B, $O_2$, revolves around $O_1$ with angular velocity $\omega$. The distance $|O_1O_2| = R_1 + R_2$. The velocity of the center of circle B is $v_{O2} = \omega (R_1 + R_2)$. The velocity of point $C$ on circle B combines its center's velocity and its rotational velocity: $v_{CB} = v_{O2} - \omega_2 R_2 = \omega (R_1 + R_2) - \omega_2 R_2$. Equating velocities at the contact point, $v_{CA} = v_{CB}$:

$$\omega_1 R_1 = \omega (R_1 + R_2) - \omega_2 R_2$$

[Q1] Solving for $\omega$:

$$\omega (R_1 + R_2) = \omega_1 R_1 + \omega_2 R_2 \implies \omega = \frac{\omega_1 R_1 + \omega_2 R_2}{R_1 + R_2}$$

[Q2] The time $t_1$ for B to circle A once is the period of revolution of the line $O_1O_2$.

$$t_1 = \frac{2\pi}{\omega} = \frac{2\pi (R_1 + R_2)}{\omega_1 R_1 + \omega_2 R_2}$$

[Q3] The time $t_2$ for B to complete one revolution relative to A is the period of this motion in A's reference frame. The relative angular velocity of the line $O_1O_2$ with respect to circle A is $\omega_{rel} = \omega - \omega_1$.

$$\omega_{rel} = \frac{\omega_1 R_1 + \omega_2 R_2}{R_1 + R_2} - \omega_1 = \frac{\omega_2 R_2 - \omega_1 R_2}{R_1 + R_2} = \frac{R_2(\omega_2 - \omega_1)}{R_1 + R_2}$$

The time is the inverse of the magnitude of this relative angular velocity:

$$t_2 = \frac{2\pi}{|\omega_{rel}|} = \frac{2\pi (R_1 + R_2)}{|R_2(\omega_2 - \omega_1)|}$$