In both cases $ AP=AQ$ implies obviosly $ \angle{ABP}=\angle{ACQ}$ and simple angle chasing shows that this, together with $ PBCQ$ being an inscribed quadrilateral, leads to $ BC'B'C$ - also cyclic. For $ I$ we get $ \angle{C'BB'}=\angle{C'CB'}$ or $ B=C$, and for $ O$ we get $ B+90-A=C+90-A$, that is $ B=C$.

for both case solutions are similar. a)draw the altitudes from $B,C$ And name them $B^*,C^*$ obviously $AO \perp B^*C^* , AO \perp PQ$ So we should have $ PQ||B^*C^*\implies BB'C'C$ cyclic. $\implies AB=AC$. b) similar to a) $\blacksquare$