a) $AI$ cuts $\Gamma$ again at the midpoint $M$ of its arc $BC.$ $OM \perp BC$ is perpendicular bisector of $\overline{BC}$ $\Longrightarrow$ $\overline{OM}$ and $\overline{IA_2}$ are congruent and parallel $\Longrightarrow$ $OMIA_2$ is a parallelogram $\Longrightarrow$ $\angle OA_1I=\angle OMI=\angle OAI$ $\Longrightarrow$ $OIAA_2$ is cyclic, in fact an isosceles trapezoid because $IA_2=AO=R$ $\Longrightarrow$ $AA_2=OI.$ Similarly $BB_2=IO.$ b) If $\tau_A$ is the reflection of $AA_2$ about $AI$ (isogonal of $AA_2$ WRT <BAC), we have then $\angle (\tau_A,AI)=\angle IAA_2=\angle OIA$ $\Longrightarrow$ $\tau_A \parallel IO$ $\Longrightarrow$ $AA_2$ contains the isogonal conjugate $U$ of the infinite point of $OI,$ which is obviously a point on $\Gamma.$ Likewise, $BB_2$ passes through $U.$

For $b)$: We see, by construction that, $A_2$ and $B_2$ are symmetrical of each other about $CI$; after having proved $AA_2=BB_2$ (excellent, as usual, by Luis), let's note $D$ the midpoint of the arc $\stackrel{\frown}{AB}$ which does not contain $C$. With the previous observations we get $\triangle AA_2D\cong\triangle BB_2D$ (s.s.s. criterion), i.e. $\widehat{AA_2, BB_2}=\widehat{ADB}$, and, indeed, $AA_2$ and $BB_2$ concur on $\Gamma$. Best regards, sunken rock

it's problem very easy: a) not difficult, $OIAA_2$ and $OIBB_2$ are isoseles trepazoid; b) $ \angle(A_2A\cap BB_2)=\angle C $;