$F$ is the homothety center of the circles $\Gamma_1$ and $\Gamma_2$. the homothety takes $I$ to $O$ and obviously we have that $OM\parallel IE$ so $E$ goes to $M$. we know that $G$ goes to $K$ - the mid-point of arc $AB$- so $MK\parallel EG$ and $MK$ is tangent to $\Gamma_1$

Let $FH$ and $FG$ cut the circle $\Gamma_{1}$ at $U$ and $R$, respectively. Easy to see that $UO\parallel IH$ and $OR\parallel IG$. The isoscales triangles $IHG$ and $UOR$ are similar, $HG\parallel UR$. Notice that as $EP = EQ$, the line $l$ is perpendicular to the angle bisector of $\angle GED$, hence $l\parallel IE$, thus $OM\perp UR$. Also notice that $\frac{IE}{OM} = \frac{FI}{FO}= \frac{HI}{UO}$. Hence $IHEG\sim OUMR$, therefore, $MR\perp OR, MR$ is the tangent from $M$ to $\Gamma_{1}$ and it is obvious that $MR\parallel AB$.

Attachments:

First of all it is a trivial observation that $IE \parallel l$.Next we see that $F,I,O$ are collinear.Let $FH$ and $FG$ meet $\Gamma_1$ at $J$ and $K$ respectively.Then $\angle{FHG}=\angle{FJK}=$ angle made by the tangent at $F$,so $HG \parallel JK$.Let $\psi$ be the homothety with center at $F$ mapping $I$ to $O$.Then $\psi:H \rightarrow J$ and $G \rightarrow K$.Also note that $\triangle{FIE} \sim \triangle{FOM}$ so $\psi:E \rightarrow M$.Thus $MK$ is tangent to $\Gamma_1$ and obviously it is parallel to $AB$.

The same solution: Let $FE$ cut $l$ at $M$. Because $IE\parallel OM$ and $F-I-O$, a homothety centered at $F$ takes $(I)$ to $(O)$, $E$ to $M$, $G$ to midpoint of arc $AB$. Call it $K$. We have $KM \parallel GE \parallel AB \implies KM$ is tangent to $\Gamma_{1}$.