1)
We have
$$\begin{aligned} \angle EGF &= \angle BGE + \angle CGF - \angle EGF \\ &= 360^{\circ} - 2\angle BAC - (180^{\circ} - 2\angle BAC) = 180^{\circ} \end{aligned}$$
then three points $E, G, F$ are collinear. Since $\angle ABK + \angle ACK = \angle AGE + \angle AGF = 180^{\circ}$ then $K$ belongs to the circle $(O)$.

It is easy to check that $G$ is the Miquel point then two triangles $GBA$ and $GKC$ are similar, which implies that $\angle BGA = \angle KGC$. We also have $GO$ is the angle bisector of $\angle BGC$ then $GO$ is the angle bisector of $\angle AGK$. Combine with $OA = OK$, we can conclude that $AOKG$ is cyclic quadrilateral. Consider the radical axis of circles $(O), (AOKG), (BOC)$, we can see that $AK, OG$ and $BC$ are concurrent.

2)
In this part, we just need $(O), (I)$ are two fixed circles that pass through $B, C$, point $D$ is fixed on $(I)$ while $G$ moves on $(I)$. By applying Pascal's theorem for the tuple $\begin{pmatrix} B & C & D \\ C & B & G \end{pmatrix}$, one can check that the line $MN$ passes through the intersection of the tangent line at $B, C$, namely $J$ of the fixed circle $(I)$. Suppose that $JD$ meet $(I)$ at the second point $X$ then $X$ is the fixed point and the quadrilateral $BCDX$ is harmonic then $G(BC, DX) = -1$. Denote $T$ as the intersection of $MN$ and $BC$ then $G(BC, DT) = -1$, which means $GX$ passes through $T$. Thus
$$\overline{TX} \cdot \overline{TG} = \overline{TB} \cdot \overline{TC} = \overline{TP} \cdot \overline{TQ}.$$
This implies that $(GPQ)$ passes through the fixed point $X$. Suppose that $Y$ is the intersection of $(GPQ)$ and $DX$ (which differs from $X$) then
$$\overline{JX} \cdot \overline{JY} = \overline{JP} \cdot \overline{JQ} = \mathcal{P}_{J/(O)}$$
which is a constant then $Y$ is fixed. Therefore, the circles $(GPQ)$ passes through two fixed points $X, Y$.