1) Denote $G$ as the intersection of $BE, CD$, then $G$ is the centroid of triangle $ABC$. We assume that $AB < AC$ and the solution is similar to all other cases.
Since $ABLC$ is cyclic, we have $180^{\circ} - \angle AKN = \angle ACN = \angle ABM = \angle AKM$, hence
$$\angle AKM + \angle AKN = 180^{\circ},$$
which means $M, K, N$ are collinear.
Since $AKNC$ and $AKBM$ are both cyclic, then by the power of a point to circle, we have
$$LN \cdot LC = LK \cdot LA = LB \cdot LM,$$
thus $MBNC$ is concyclic and we can see $\angle CBL = \angle LNM$.
Denote $P$ as the intersection of $OL$ and $MN$. Since $O$ is the circumcenter of $(ABLC)$ then
$$\angle PLN = 90^{\circ} - \frac{\angle LOC}{2} = 90^{\circ} - \angle CBL = 90^{\circ} - \angle MNL$$
Therefore, $OL$ is perpendicular to $MN$.

2) Notice that $ADGEB C$ is a complete quadrilateral and $K$ is the Miquel point so $K$ belongs to two circles ($BDG$), ($CEG$). Thus we have
$$\angle DKB = \angle DGB = \angle EGC = \angle EKC, \quad \angle KDB = \angle KGB = \angle KCE.$$
So $\triangle BKD \sim \triangle EKC$, then by sin law, we have
$$\frac{AB}{AC} = \frac{BD}{CE} = \frac{BK}{KE} = \frac{\sin BAK}{\sin EAK} = \frac{\sin BAL}{\sin CAL}$$
which means $ABLC$ is the harmonic quadrilateral and $LA$ is the symmedian of triangle $LBC$.
Finally, because $MN$ is the antiparallel to $BC$ with respect to $\angle BAC$ then the symmedian of triangle $LBC$ is the median of triangle $LMN$, which means $LA$ passes through the midpoint of $MN$ or $K$ is the midpoint of the segment $MN$. $\square$