1) We use inversion to solve this problem.

Suppose that $AI$, $A_{1}I$ intersect $(O)$ at $A_{0}$, $A_{1}$. Because $AI$ is the bisector then $A_{0}$ is the midpoint of the minor $\operatorname{arc} BC$. Based on the property of the Mixtilinear circle, we also have $A_{1}$ as the midpoint of the major arc $BC$.
Consider the inversion of center $I$ and ratio equal to the power of $I$ to $(O)$ as the function $f$.
We have $f(A) = A_{0}$, $f(A_{1}) = A_{2}$ then $f(AA_{1}) = (IA_{0}A_{2})$. Define $B_{0}$, $B_{2}$, $C_{0}$, $C_{2}$ similarly then $f(BB_{1}) = (IB_{0}B_{2})$, $f(CC_{1}) = (IC_{0}C_{2})$.
It is easy to see that three circles $(IA_{0}A_{2})$, $(IB_{0}B_{2})$ and $(IC_{0}C_{2})$ share the common point $I$. On the other hand, the power of $O$ to the three circles is also equal to $-R^{2}$ where $R$ is the radius of the circumcircle.
Hence, the three circles have two common points and one of them is $I$ which is the center of inversion. Then $AA_{1}$, $BB_{1}$, $CC_{1}$ are concurrent at a point $M$ and $M$, $I$, $O$ are collinear.

2) From the property of the Mixtilinear circle, we have $B_{a}B_{c}$ and $C_{a}C_{b}$ have the common midpoint $I$ then $B_{a}C_{a}B_{c}C_{b}$ is a parallelogram, which implies that $B_{a}C_{a}$ is parallel to $BC$ and the distance from $I$ to $B_{a}C_{a}$ and $BC$ are the same. Hence $B_{a}C_{a}$ is tangent to $(I)$. Similarly, we also have $A_{b}C_{b}$ and $B_{c}A_{c}$ are also tangent to $(I)$.

It is easy to see that $C_{b}B_{c}$ coincides with $BC$ then it is tangent to $(I)$. Similarly, we also have $C_{a}A_{c}$ and $A_{b}B_{a}$ are tangent to $(I)$.
Hence, the 6 sides of the hexagon $C_{b}B_{c}A_{c}C_{a}B_{a}A_{b}$ are tangent to the circle $(I)$, which finishes the solution.