a) The angles $\angle ABC$ and $\angle AB_1C_1$ are equal, therefore so are their complementary angles, $\angle BAA_1$ and $\angle A_2AC$.
It follows that the rays ($AH$ and ($AA_2$ are isogonal, hence $A_2 \in (AO$). As $AO = CO$, we have $\angle ACO \equiv \angle OAC \equiv \angle AA_2B_1$, therefore points $O, A_2, B_1, C$ are cocyclic. (The arguments above hold in the case $A_2 \in (AO)$ as well as in the case $O \in (AA_2)$.)

b) From the power of the point $A$ with respect to the circles through $O, A_2, B_1, C$ and $H, A_1, C, B_1$ respectively, it follows that $AB_1 \cdot AC = AO \cdot AA_2$ and $AB_1 \cdot AC = AH \cdot AA_1$.
Since $AO \cdot AA_2 = AH \cdot AA_1$, from the converse of the power of the point theorem, it follows that the points $O, A_2, H, A_1$ are cocyclic.

Alternative Solution. For b). Consider $\{A_3\} = (AA_2 \cap B_1C_1$ and let $O_1$ be the midpoint of $[AH]$. Then $O_1$ is the circumcenter of triangle $AB_1C_1$. As $[O_1A_3]$ is a midsegment in triangle $AHA_2$, we have $\angle HA_2A \equiv \angle O_1A_3A$. But the similarity of triangles $ABC$ and $AB_1C_1$ leads to the equality of the corresponding angles $\angle AA_1O$ and $\angle AA_3O_1$, which means that $\angle AA_2H \equiv \angle AA_1O$ and the conclusion.