I know a few methods referred to as "pure geometry," but I still want to demonstrate the usefulness of complex numbers in this problem 1) We notice that $H$ lies on the line segments $AD, BE, CF$, which are chords of the circles $(AXD), (BYE), (CZF)$. Therefore, $H$ has the same power with respect to these circles, and this power is evidently negative. Hence, to prove that $(AXD), (BYE), (CZF)$ each have two common points with $OH$, it suffices to show that the line joining the centers of $(BYE)$ and $(CZF)$ is perpendicular to $OH$. Similarly, if the line joining the centers of $(AXD)$ and $(BYE)$ is perpendicular to $OH$, the proof will be complete. 2) We also observe that $NP, PM, MN$ are the perpendicular bisectors of $AD, BE, CF$. From this, $XX', YY', ZZ'$ are the diameters of $(AXD), (BYE), (CZF)$. Therefore, it suffices to prove $Y'Z' \perp HJ$, and similarly $Z'X' \perp HJ$, which will solve the problem. In the complex plane, let $(O)$ be the unit circle with $A(1), B(b^2), C(c^2)$. Then the orthocenter $H$ has coordinates $h = 1 + b^2 + c^2$ and $i = -(bc + c + b)$. From the midpoints $M, N, P$ of $BC, CA, AB$, we find \[ m = \frac{b^2 + c^2}{2}, \, n = \frac{c^2 + 1}{2}, \, p = \frac{b^2 + 1}{2}. \]From $E, F$, the intersections of $HB, HC$ with $CA, AB$, respectively, we obtain \[ e = \frac{b^4 + b^2c^2 + b^2 - c^2}{2b^2}, \, f = \frac{c^4 + b^2c^2 - b^2 + c^2}{2c^2}. \]From $E, F$, the intersections of $IB, IC$ with $MP, MN$, respectively, we obtain \[ y = \frac{c^2b^2 - c^2 + 2cb^2 + b^4 + b^2}{2(c + b^2)}, \, z = \frac{b^2c^2 - b^2 + 2bc^2 + c^4 + c^2}{2(b + c^2)}. \]From this, with $K, L$ being the centers of $(BYE), (CZF)$, respectively, we find \[ k = \frac{b^2(c^2b^2 - 3c^2 + b^4 + b^2)}{2(b^2 - c)(c + b^2)}, \, l = \frac{c^2(b^2c^2 - 3b^2 + c^4 + c^2)}{2(c^2 - b)(b + c^2)}. \]Thus, the ratio \[ \frac{k - l}{h} = \frac{(b - c)^3(b + c)^3}{2(b - c^2)(b + c^2)(b^2 - c)(b^2 + c)}. \]It is evident that this is a purely imaginary number because replacing $b, c$ with $\bar{b} = \frac{1}{b}, \bar{c} = \frac{1}{c}$ gives its negative. Hence, $YZ \perp OH$. This concludes the proof. We note that $Y', Z'$ are the reflections of $Y, Z$ over $K, L$, respectively. Therefore, \[ y' = 2k - y = \frac{b^4 + b^2c^2 - 2b^2c + b^2 - c^2}{2b^2 - 2c}, \, z' = 2l - z = \frac{-c^4 - b^2c^2 + 2bc^2 + b^2 - c^2}{-2c^2 + 2b}. \]Moreover, $j = -i = bc + b + c$. We compute the ratio \[ \frac{y' - z'}{h - j} = \frac{(b - c)(b + c)^2}{2(b - c^2)(b^2 - c)}. \]It is also evident that this is a purely imaginary number (as replacing $b, c$ with $\bar{b} = \frac{1}{b}, \bar{c} = \frac{1}{c}$ gives its negative). Thus, $Y'Z' \perp JH$. This concludes the proof.

This is my solution for this problem: a) Let $A=(1,0,0), B=(0,1,0), C=(0,0,1)$. We have: $$N=(1,0,1)$$$$P=(1,1,0)$$$$I=(a,b,c)$$The equation of line $NP$ is $$-x+y+z=0$$The equation of line $AI$ is $$cy-bz=0$$Then we have $$X=(b+c,b,c)$$We have $$H=(S_{B}S_{C},S_{A}S_{C},S_{A}S_{B})$$$$D=(0,S_{C},S_{B})=(0,a^2+b^2-c^2,a^2+c^2-b^2)$$The equation of circle $(AXD)$ is $$-a^{2}yz-b^{2}xz-c^{2}xy+(u_{A}x+v_{A}y+w_{A}z)(x+y+z)=0$$Put $A$ in the equation, we have $u_{A}=0$ Put $X$ in the equation, we have: $$v_{A}b+w_{A}c=\frac{a^{2}bc+bc(b+c)^2}{2(b+c)}$$Put $D$ in the eqution, we have: $$v_{A}(a^2+b^2-c^2)+w_{B}(a^2+c^2-b^2)=\frac{a^4-(b^2-c^2)^2}{2}$$This give us $$v_{A}=\frac{c^2(a^2+c^2-b^2)}{2(c^2-b^2)}, w_{A}=\frac{b^2(a^2+b^2-c^2)}{2(b^2-c^2)}$$Similarly, we have the equation of circle $(BYE)$ is $$-a^{2}yz-b^{2}xz-c^{2}xy+(u_{B}x+v_{B}y+w_{B}z)(x+y+z)=0$$with $$u_{B}=\frac{c^2(b^2+c^2-a^2)}{2(c^2-a^2)}, v_{B}=0, w_{B}=\frac{a^2(a^2+b^2-c^2)}{2(a^2-c^2)}$$We have the equation of the radical axis of $(AXD)$ and $(BYE)$: $$(u_{A}-u_{B})x+(v_{A}-v_{B})y+(w_{A}-w_{B})z=0$$$$(b^2+c^2-a^2)(b^2-c^2)x+(a^2+c^2-b^2)(c^2-a^2)y+(a^2+b^2-c^2)(a^2-b^2)z=0$$But this is also the equation of line $OH$, and notice that $H$ is inside all three circle give us the solution of a). b) Let $(ABC)$ be the unit circle. We have $$h=a^2+b^2+c^2$$$$i=-ab-bc-ca$$$$j=2o-i=ab+bc+ca$$Let $X_{*}$ be the intersection of $AI$ with $(O)$. Let $X_{1}$ be a point on line $BC$ such that $AX_{1}$ be the external bisector of point $A$ wrt $\triangle BAC$ We have $$\overline{x_1}=\frac{b^2+c^2-x_1}{b^{2}c^{2}}, \frac{x_{1}-a^2}{\overline{x_{1}}-\frac{1}{a^2}}=-\frac{a^2-x_{*}}{\overline{a^2}-\overline{x_{*}}}$$Therefore $$x_{1}=\frac{a^2(b^2-bc+c^2)-b^{2}c^{2}}{a^2-bc}$$So $$x'=\frac{a^2+x_{1}}{2}=\frac{a^4+a^2(b^2-2bc+c^2)-b^2c^2}{2(a^2-bc)}$$Similarly, we have $$y'=\frac{a^2(b^2-c^2)-2b^2ca+b^4+b^2c^2}{2(b^2-ac)}$$$$z'=\frac{a^2(c^2-b^2)-2c^2ba+c^4+b^2c^2}{2(c^2-ab)}$$Therefore $$y'-z'=\frac{(b+c)^2(c-b)(a^2+b^2+c^2-ab-bc-ca)}{2(b^2-ac)(c^2-ab)}$$So $$\frac{y'-z'}{h-j}=-\overline{\left(\frac{y'-z'}{h-j}\right)}$$We then have $Y'Z' \perp HJ$, similar calculation give us the solution of part b)

Here are the aforementioned "pure geometry" methods

Here is my solution for this problem Solution a) Let $D_a \equiv AI \cap BC,$ $X_a \in BC$ satisfies $AX_a$ be tangent at $A$ of $(ABC)$. Note that $(X_a, X_aA)$ is $A$ - Apollonius circle of $\triangle ABC$ then $D_a \in (X_a, X_aA)$. Combine with $X$ is midpoint of $AD_a,$ we have $AX_a$ is diameter of $(AXD)$. From this, we have $\mathcal{P}_{O / (AXD)} = OA^2 = R^2_{(O)},$ with $R_{(O)}$ is radius of $(ABC)$. Similarly, we have $$\mathcal{P}_{O / (AXD)} = \mathcal{P}_{O / (BYE)} = \mathcal{P}_{O / (CZF)} = R^2_{(O)} > 0$$On the other side, we have $\mathcal{P}_{H / (AXD)} = \overline{HA} \cdot \overline{HD} = \dfrac{1}{2} \mathcal{P}_{H / (ABC)}$. Similarly, we have $$\mathcal{P}_{H / (AXD)} = \mathcal{P}_{H / (BYE)} = \mathcal{P}_{H / (CZF)} = \dfrac{1}{2} \mathcal{P}_{H / (ABC)} < 0$$Therefore, $(AXD), (BYE), (CZF)$ have 2 common points lie on $OH$. b) We have $$\angle{XAX'} = \angle{XAD} + \angle{DAX'} = \angle{XAD} + \angle{DXX'} = \angle{XAD} + \angle{AXX'} = 90^{\circ}$$Then $AX'$ is external angle bisector of $\angle{BAC}$. Let $D'_a \equiv AX' \cap BC;$ $I_a, I_b, I_c$ be excenter of $\triangle ABC$. It's easy to see that $J$ is center of $(I_aI_bI_c)$. Since $(I_bI_c, AD'_a) = - 1$ and $X'$ is midpoint of $AD'_a,$ we have $X'A^2 = \overline{X'I_b} \cdot \overline{X'I_c} = \mathcal{P}_{X' / (I_aI_bI_c)}$. This means $(X', X'A)$ and $(I_aI_bI_c)$ are orthogonal. Hence $\mathcal{P}_{J / (X', X'A)} = R^2_{(I_aI_bI_c)},$ with $R_{(I_aI_bI_c)}$ is radius of $(I_aI_bI_c)$. Similarly, we have $$\mathcal{P}_{J / (X', X'A)} = \mathcal{P}_{J / (Y', Y'B)} = \mathcal{P}_{J / (Z', Z'C)} = R^2_{(I_aI_bI_c)}$$We also have $\mathcal{P}_{H / (X', X'A)} = \overline{HA} \cdot \overline{HD} = \dfrac{1}{2} \mathcal{P}_{H / (ABC)}$. Similarly, we have $$\mathcal{P}_{H / (X', X'A)} = \mathcal{P}_{H / (Y', Y'B)} = \mathcal{P}_{H / (Z', Z'C)} = \dfrac{1}{2} \mathcal{P}_{H / (ABC)}$$Then $JH$ is common radical axis of $(X', X'A); (Y', Y'B); (Z', Z'C)$ of $\overline{X', Y', Z'} \perp JH$.

I solved part (a), but got stuck on part (b) We show that the power of $H$ and $O$ with respect to the three circles is equal. It is a well known fact that $AH \cdot HD=BH \cdot HE=CH\cdot HF=4R^2 \cos A \cos B\cos C$ where $R$ is the circumradius of $ABC$. Next, we show that $OA$ is tangent to $(AXD)$. Indeed, note that $PN$ is the perpendicular bisector $AD$, so $\angle XDA=\angle XAD=\angle \frac{B-C}{2}=\angle XAO$, so $Pow_{(AXD)}(O)=R^2$, which is constant.