1)
Since $EF$ and $I_bI_c$ are both perpendicular to $AI$ then $I_bI_cFE$ is the trapezoid. So $PQ$ is the midline of both trapezoid $I_bI_cFE$ and triangle $AEF$. Thus $P$, $Q$ belong to the radical axis of the degenerate circle ($A$, $0$) and $(I)$. Similarly, $Q$ belongs to the radical axis of ($B$, $0$) and $(I)$. Hence, $QA^2 = \overline{QF} \cdot \overline{QY} = QB^2$, which implies that $(QAB)$ is tangent to $(I)$ at $Y$. Similarly, $(PAC)$ is also tangent to $(I)$ at $X$.

Thus, $(I)$ is the S-Mixtilinear of triangle $ASB$ then the incenter of triangle $ABS$ is the midpoint $N$ of the segment $EF$ which implies that $SQ$ is the angle bisector of $\angle ASB$ and then $SQ$ passes through $N$. Similarly, $RP$ also passes through $N$. Therefore, $PR$, $QS$, $AI$ are concurrent at point $N$.

2)
In the circle $(I)$, the line $I_bI_c$ is the antipole of $N$ then $JE$, $KF$, $DN$ are concurrent at point $M$ that lies on circle $(I)$. We have the following theorem: (Steinbart's theorem) Let $ABC$ be a triangle with $(I)$ and this circle is tangent to $BC$, $CA$, $AB$ at $D$, $E$, $F$ respectively. Take three points $X$, $Y$, $Z$ on the circle $(I)$, then $AX$, $BY$, $CZ$ are concurrent if and only if $DX$, $EY$, $FZ$ are concurrent.

By applying this, we can see that to prove three lines $AM$, $BY$, $CZ$ are concurrent, we have to prove that $FX$, $EY$, $DM$ are concurrent.

We have $\angle EYF = \angle AEF = \angle I_bAC$ then quadrilateral $I_cAEY$ is cyclic. Similarly, the quadrilateral $AI_bXF$ is also cyclic. These mean points $X$, $Y$ defined above are the same as definition in problem.

Consider the transformation $S$ which is the union between the inversion $I_A^{AB \cdot AC}$ and the reflection respect to the line $R_{AI}$. We have $S : (O) \leftrightarrow BC$, $(I) \leftrightarrow (T)$ with $(T)$ is the ex-mixtilinear respect to vertex $A$ of triangle $ABC$. This circle is tangent to $AC$, $AB$ at $E'$, $F'$ respectively then $E \leftrightarrow E'$, $F \leftrightarrow F'$.

From the Sawayama's lemma, the excenter $I_a$ is the midpoint of segment $E'F'$. By applying the Pappus's theorem for two tuples $(I_c, A, I_b)$ and $(E', I_a, F')$, we have $I_bE'$ meets $I_cF'$ at point $Z$ which belongs to $BC$.

Denote $G$ as the tangent point of $(T)$ with $(O)$. We already know that $I_aG$ passes through point $L$, the midpoint of the arc $BAC$ of circle $(O)$ which is also the midpoint of $I_bI_c$. But $I_bI_c \parallel E'F'$ then since Thales's theorem, we have $L$, $Z$, $G$, $I_a$ are collinear.

We have $S: E'I_b \leftrightarrow (I_cAE)$, $F'I_c \leftrightarrow (I_bAF)$, $D \leftrightarrow G$, $I \leftrightarrow I_a$ then $GI_a \leftrightarrow (AID)$. Since $E'I_b$, $F'I_c$, $I_aG$ are concurrent then circles
$$(AI_cE), (AI_bF), (AID)$$
are coaxial. We have $\overline{NM} \cdot \overline{ND} = \overline{NE} \cdot \overline{NF} = \overline{NA} \cdot \overline{NI}$ then $AMID$ is cyclic. Consider the radical axis of three circles $(I)$, $(I_cAE)$, $(I_bAF)$, we have $EY$ cuts $FX$ at $U$ which is the radical center of these circles. Continue to consider the radical axis of three circles $(I)$, $(I_cAE)$, $(AID)$, we have $MD$ cuts $EY$ at $U'$ which is the radical center of these circles. But $(AI_cE)$, $(AI_bF)$, $(AID)$ are coaxial, which implies that $U \equiv U'$.

Therefore, three lines $MD$, $EY$, $FX$ are concurrent at $U$. The problem is solved completely. ■