Denote $F$ as the intersection of $AD$, $BC$. Suppose that $FK$ intersects $AB$, $CD$ at $S$, $T$ respectively. Since the harmonic points of complete quadrilateral, we have $(E, T, D, C) = -1$. Then $ET \cdot EH = ED \cdot EC$. Similarly, $ES \cdot EG = EA \cdot EB$. But $EA \cdot EB = ED \cdot EC$ then $ES \cdot EG = ET \cdot EH$ which implies that $G, H, T, S$ are concyclic.

Denote $Q'$ as the midpoint of $FK$ then $Q'$, $G$, $H$ belong to the Gauss line of the complete quadrilateral $AFBKCD$. We shall prove that $Q'$ belongs to the radical axis of $(O), (I)$.
It is easy to see that $EK$ is the antipole of $F$ then denote $EK \cap (O) = \{X, Y\}$ then $FX$, $FY$ are the tangent lines of $(O)$. Let $U, V$ be midpoints of $FX$, $FY$.
We consider the power of point to circle $(O)$ and degenerate circle $F$.
$$\mathscr{P}_{U/(O)} = UX^2, \quad \mathscr{P}_{U/(F)} = UF^2$$
but $UX = UF$ then $\mathscr{P}_{U/(O)} = \mathscr{P}_{U/(F)}$. This implies that $U$ belongs to the radical axis of $(O), (F)$. Similarly with the point $V$ then $UV$ is the radical axis of $(O), (F)$. But $Q' \in UV$ which is the midline of triangle $FXY$, then
$$\mathscr{P}_{Q'/(F)} = \mathscr{P}_{Q'/(O)} \Leftrightarrow \mathscr{P}_{Q'/(O)} = Q'F^2 = Q'K^2 = Q'S \cdot Q'T = Q'G \cdot Q'H = \mathscr{P}_{Q'/(I)}.$$
Then $Q'$ belongs to the radical axis of $(I), (O)$ which means $Q' \in MN$ or $Q' \equiv Q$.
Thus $QK^2 = \mathscr{P}_{Q/(I)}$ implies that $QK$ is the tangent line of $(I)$. Hence, $QK \perp KI$.
Using the Brocard's theorem, we have $K$ is the orthocenter of triangle $OEF$ then $EO \perp FK$. Combining all these results, we get $EO \parallel IK$.

2) Two lines pass through $Q$ and intersect $(I)$ at $M, N$ and $G, H$ then $P$ is the intersection of $MG$, $NH$ which means $P$ belongs to the antipole of $Q$.
Note that $QK$ is tangent to $(I)$ which means $K$ also belong to the antipole of the pole $Q$ respect to circle $(I)$.
Then $PK$ is the antipole of $Q$ respect to the circle $(I)$. This implies that $PK \perp IQ$.