Connect AC, then AC will pass through point O and extend FO from AB to M;

∵OF⊥CD, OE⊥BC, quadrilateral ABCD is a square,

∴ Quadrilateral OECF is a square,

∴OM=OF=OE=AM，

∠∠ Mao =∠OFE=45, ∠AMO=∠EOF=90,

∴△AMO≌△FOE，

∴AO=EF, and ∠ aom = ∠ ofe = ∠ foc = 45, which means OC⊥EF.

Therefore, AP=EF, and AP ⊥ ef.

(2) The conclusion of question (1) is still valid for the following reasons:

Extend AP to BC in N and FP to AB in M;

The MBE of pe⊥bc ∵pm⊥ab is 90, MBP = ∠ EBP = 45,

The quadrilateral MBEP is a square,

∴mp=pe,∠amp=∠fpe=90；

∫a b-BM = AM，BC-BE=EC=PF，AB=BC，BM=BE，

∴AM=PF，

∴△AMP≌△FPE，

∴AP=EF,∠APM=∠FPN=∠PEF

∠∠PEF+∠PFE = 90，∠FPN=∠PEF，

∴∠ fpn+∠ pfe = 90, that is, AP⊥EF.

Therefore, AP=EF, and AP ⊥ ef.

(3) (1) The conclusion of (2) is still valid;

As shown on the right, extend AB to PF at H, which is exactly the same as (2).