Solution:
1. Let $AP \perp FG$ and $BQ \perp FG$, $P, Q \in FG$. Then
$$AB \geq PQ = PG + GF + FQ$$
Since the quadrilateral $CFMG$ is cyclic we obtain $\angle CMF = \angle CGF = \angle AGP$. This implies that $\triangle APG \sim \triangle CFM$ and therefore
$$\frac{PG}{AG} = \frac{MF}{CM} \Longleftrightarrow PG = \frac{MF \cdot AG}{CM}$$

In the same way we have $QF = \frac{MG \cdot BF}{CM}$. Then
$$AB - FG \geq PG + FQ = \frac{MF \cdot AG + MG \cdot BF}{CM}$$
It is clear that equality is attained iff $AB \parallel FG$. The last means that $\angle BAC = \angle FGC = \angle AGP$, whence $\angle MCB = 90^\circ - \angle BAC = \angle OCB$, where $O$ is the circumcenter of $\triangle ABC$. Therefore the required locus is the segment $CD$, where $D$ is the intersection point of the line $OC$ and the side $AB$.