a) Let the line passing through $A$ and parallel to $BC$ meet $(O)$ at $A_0$. We have $AA_0CB$ is an isosceles trapezoid so $A_0C = AB = 2FH_a$. We also have $\angle FH_aB = \angle FBH_a = \angle A_0CB$ then $FH_a \parallel A_0C$. But $GC = 2GC$, it follows that $G$, $A_0$ and $H_a$ are collinear.

Let $S$ be the midpoint of $A_0H_a$ then we have $SE \parallel AA_0$ so $\angle AES = \angle A_0AC = \angle A_0XC$, which implies $SECX$ is a cyclic quadrilateral.

Let $T$ be the midpoint of $BH_a$ then we have $ST \parallel BA_0$. We obtain $\angle XST = \angle XA_0B = \angle XCB$ and it follows that $XCST$ is a cyclic quadrilateral.

From the above, we have $X, C, E, S$ and $T$ lie on the same circle, thus the circumcircle of triangle $XEC$ passes through the midpoint of $BH_a$.

b) Firstly, we will prove the following lemma.

Lemma. Given triangle $ABC$ with circumcircle $(O)$ and $P, Q$ are two arbitrary points. Lines $PA, PB$ and $PC$ intersect $(O)$ at the second points $P_a, P_b$ and $P_c$. Lines $QA, QB$ and $QC$ intersect $(O)$ at the second points $Q_a, Q_b$ and $Q_c$. Lines $P_aQ_a, P_bQ_b$ and $P_cQ_c$ intersect $BC, CA$ and $AB$ at $D, E$ and $F$ respectively. Then the points $D, E$ and $F$ are collinear.

Proof. Using the property of cyclic quadrilaterals, we have
$$\frac{DB}{DC} = \frac{P_aB}{P_aC} \cdot \frac{Q_aB}{Q_aC}.$$
From there, we obtain that
$$\begin{aligned} \prod \frac{DB}{DC} &= \prod \left( \frac{P_aB}{P_aC} \cdot \frac{Q_aB}{Q_aC} \right) = \prod \frac{P_aB}{P_aC} \cdot \prod \frac{Q_aB}{Q_aC} \\ &= \prod \frac{\sin PAB}{\sin PAC} \cdot \prod \frac{\sin QAB}{\sin QAC} = 1. \end{aligned}$$
Using Menelaus' theorem, we deduce that three points $D, E$ and $F$ are collinear. $\square$

Back to the problem, according to part a), $AA_0 \parallel BC$ so $XA$ and $XA_0$ are isogonal with respect to $\angle BXC$. Let $U$ be the intersection of $AX$ and $BC$. We have
$$\frac{UB}{UC} \cdot \frac{H_aB}{H_aC} = \frac{XB^2}{XC^2}. \qquad (1)$$
It is also well-known that
$$\frac{UB}{UC} = \frac{BA}{CA} \cdot \frac{BX}{XC}.$$
Therefore,
$$\frac{XB^2}{XC^2} = \frac{UB^2}{UC^2} \div \frac{AB^2}{AC^2}.$$
Combining with (1), we get
$$\frac{UB}{UC} = \frac{UB^2}{UC^2} \div \frac{AB^2}{AC^2} \div \frac{H_aB}{H_aC}.$$
Hence,
$$\frac{UB}{UC} = \frac{AB^2}{AC^2} \cdot \frac{H_aB}{H_aC}.$$
Let $V$ be the intersection of $BY$ and $CA$, $W$ be the intersection of $CZ$ and $AB$ then we have similar equality. Thus,
$$\prod \frac{UB}{UC} = \prod \left( \frac{AB^2}{AC^2} \cdot \frac{H_aB}{H_aC} \right) = 1.$$
It follows that $AX$, $BY$, and $CZ$ are concurrent at $I$.

Let $A_1$, $B_1$ and $C_1$ be the symmetric points of $A$, $B$ and $C$, respectively, through the midpoints of $BC$, $CA$ and $AB$. By simple calculation, we observe that $A_1X \parallel DM$ and $\frac{GA_1}{GD} = 3$.

Hence the homothety with center $G$ in ratio $3$ turns the line $DM$ into a straight line $A_1X$. Similarly, we can prove that this homothety also transforms $EN$ and $FP$ to $B_1Y$ and $C_1Z$.

From here, we just need to prove that the lines $A_1X$, $B_1Y$ and $C_1Z$ are concurrent. Consider triangle $XYZ$ and two points $I$ and $G$, we have $XI$, $YI$ and $ZI$ intersect $(XYZ)$ at $A$, $B$ and $C$; $XG$, $YG$ and $ZG$ cut $(XYZ)$ at $A_0$, $B_0$ and $C_0$. Applying the lemma, we have $YZ$, $ZX$ and $XY$ cut $AA_0$, $BB_0$ and $CC_0$ in three collinear points. Applying Desargues' theorem, we conclude that $A_1X$, $B_1Y$ and $C_1Z$ are concurrent. $\square$