a) The triangle $ADC$ is isosceles ($AD = AC$ are radius of $\Gamma_2$), so $\angle D_1 = \angle C$.
The angle $\angle F_1$ is external of the cyclic $ACBF$, therefore $\angle F_1 = \angle C$.
From the two equalities above we conclude that $\angle D_1 = \angle F_1$, thus
$$BD = BF \tag{1}.$$

The angle $\angle D_2$ is the half of the center angle $E\widehat{A}C$, so
$$\angle D_2 = \frac{E\widehat{A}C}{2} \quad (a).$$
Moreover the angles $\angle B_1$ and $E\widehat{A}C$ see the arc $EC$ in $\Gamma_1$, so
$$\angle B_1 = E\widehat{A}C \quad (b)$$
From the triangle $BDE$ we have:
$$\angle E_1 = \angle B_1 - \angle D_2 \stackrel{(a),(b)}{=} E\widehat{A}C - \frac{E\widehat{A}C}{2} = \frac{E\widehat{A}C}{2}$$
$$\text{Therefore } \angle D_2 = \angle E_1, \text{ so } BD = BE \quad (2).$$
From (1) and (2) we conclude that $B$ is the center of $\Gamma_3$.

b) The bisector of $\angle B_1$ is perpendicular bisector of $EG$ and passes through the midpoint (let it be $M$) of the arc $CE$.
We also have the equality $OC = OE$ (both are radius of $\Gamma_1$) and $AC = AE$ (both are radius of $\Gamma_2$). Therefore, the line $OA$ is perpendicular bisector of $CE$, thus it passes through the midpoint $M$ of the arc $CE$.
We conclude that $M$ of the arc $CE$ is the circumcenter of $CEG$, as point of intersection of the perpendicular bisectors of $EG$ and $CE$.
Therefore $CA \perp CM$, so $CA$ is tangent to the circumcircle of $CEG$.