Are you sure you posted the problem correctly?

tudor129 wrote: Are you sure you posted the problem correctly? There is one mistake, thank you

My drawing?

tudor129 wrote: My drawing? The corrected one is "Let $D, G$ be reflections to $A$ over $O$ and circumcentre of $(AEF)$, respectively", not "Let $D, G$ be reflections to $A$ and $O$ over circumcentre of $(AEF)$" Very apologize to that typo

Here's my solution for $\texttt{(a)}$ Let $AG\cap FE = Y$ and $R$ denote the circumradius of $\triangle ABC$. Note that $GA$ touches $(O)$ at $A$, and $$\frac{AY}{GY} = \frac{R}{\cot(C-B)}\cdot \frac{\cot(C-B)}{R\cdot \tan(B)\cdot \tan(C)} = \cot(B)\cdot \cot(C)\hspace{0.2 cm} \wedge \hspace{0.2 cm} \frac{YO}{KO} = \frac{\cos(A)}{\cos(C-B)}$$So $$\frac{AG}{GY}\cdot \frac{YK}{KO}\cdot \frac{OD}{DA} = \left(1+\frac{AY}{GY}\right)\left(1+\frac{YO}{KO}\right)\cdot \frac{1}{2} = \frac{1 + \tan(B)\cdot \tan(C)}{\tan(B)\cdot \tan(C)} \cdot \frac{\cos(A)\cdot \cos(C-B)}{\cos(C-B)} \cdot \frac{1}{2} = 1$$and we're done by Menelaus' Thm.

Synthetic solution for $(a)$ Let $R=(ABC)\cap (AEF)$. Obviously $G-R-D$ are collinear. Since $\angle BEO=\angle OAC$ we get $OA$ tangents to $(AEF)$. So we get $OR$ also tangents to $(AEF)$, which means $AREF$ is harmonic and projecting this to $(ABC)$ through $A$ gives that $BCDR$ is harrmoic, which means $K-D-R$ are colliner. So we get $K-D-R-G$ are collinear, which finish part $(a)$.

a) Let $S$ be second intersection of $(O)$ and $(AEF)$. Then it's easy to see that $D, S, G$ are collinear. But $(AO, AF) \equiv \dfrac{\pi}{2} - (BC, BA) \equiv (EA, EF) \pmod \pi$ so $\triangle OAF$ $\stackrel{+}{\sim}$ $\triangle OEA$ or $OA^2 = \overline{OE} \cdot \overline{OF},$ which means $OA$ tangents $(AEF)$ at $A$. Hence $A(BC, SD) = A(EF, SO) = - 1,$ or $SBDC$ is harmonic quadrilateral, then $S, D, K$ are collinear, so $S, G, D, K$ are collinear b) Redefine $H$ is intersection of the line through $I$ that perpendicular to $AO$ and perpendicular bisector of $IK;$ $P, Q$ be intersection of $IH$ with $AB, CA;$ $HK, AO$ intersect $BC$ at $W, U$; $V \in BC$ such as $(BC, UV) = - 1$. We have $(PQ, PA) \equiv \dfrac{\pi}{2} - (AB, AO) \equiv (CA, CB) \pmod \pi$. Then $B, C, P, Q$ lie on a circle. So $\mathcal{P}_{I / (O)} = \overline{IB} \cdot \overline{IC} = \overline{IP} \cdot \overline{IQ} = \mathcal{P}_{I / (APQ)}$ or $I$ lies on radical axis of $(O)$ and $(APQ)$. We also have $(KI, KW) \equiv (IH, IK) \equiv (UA, UB) \pmod \pi,$ then $O, U, K, W$ lie on a circle. Hence $$\overline{IU} \cdot \overline{IW} = \overline{IO} \cdot \overline{IK} = - IB^2 = - \overline{IU} \cdot \overline{IV}$$or $I$ is midpoint of $VW$. Therefore $$I(MN, HK) = A(CB, O \infty) = \dfrac{\overline{UC}}{\overline{UB}} = \dfrac{\overline{VC}}{\overline{VB}} = \dfrac{\overline{WB}}{\overline{WC}} = K(BC, WI) = K(MN, HI)$$which means $H, M, N$ are collinear