1. Take the midpoint of BC as point D, and connect PD and AD. Because AB = AC and Pb = PC, according to the properties of isosceles triangle, PD⊥BC,AD⊥BC and PD intersect with AD at point D in the plane APD, so BC⊥ plane APD, so PA ⊥.

2. In a planar PCB, parallel lines of a straight line BC pass through point E, point PB O, and then connect to FO.

From PE: EC = AF: FB = 3: 2, we can get PE: EC = AF: FB = PO: OB, so the straight line FO||AP.

In the triangle FEO, the angle OFE is the angle α formed by AP and EF, and the angle OEF is the angle β formed by BC and EF.

BC⊥AP is proved from the above problems, and then EO⊥FO can be deduced, so the triangle of EOF is a right triangle, so α+β = 90.