∵ quadrilateral ABCD is a square,
∴CD=CB，
∴CF= 12BC，
∫CB′= CB，
∴cf= 12cb′
∴, sin∠CB'f = cfcb'= 12 in rt△b'fc,
∴∠cb′f=30，
(2) As shown in Figure 2, connect the intersection CG of BB' to point K, and it can be seen from the folding that EF bisects AB vertically.
∴b′a=b′b，
∠B′AE =∠B′BE，
∵ quadrilateral ABCD is a square,
∴∠ABC=90，
∴∠b′be+∠kbc=90，
According to the folding, ∠ bkc = 90,
∴∠KBC+∠GCB=90，
∴∠b′be=∠gcb，
We also know from the folding that ∠GCB=∠GCB',
∴∠b′ae=∠gcb′，
(3) the quadrilateral B'PD'q is a square,
Proof: As shown in Figure 3, connect AB'
It can be seen from (2) that ∠ b ′ AE = ∠ gcb ′, from folding, ∠ gcb ′ = ∠ PCN,
∴∠b′ae=∠pcn，
According to the folding, ∠ AEB' = ∠ CNP = 90, AE= 12AB, CN= 12BC,
Similarly, the quadrilateral ABCD is a square,
∴AB=BC，
∴AE=CN，
At △AEB' and △CNP.
∠B′AE =∠PCNA e = CN∠AEB′=∠CNP
∴△aeb′≌△cnp(asa)
∴eb′=np，
Similarly, EB'=MQ,
According to symmetry, EB ′ = FD ′,
∴eb′=np=fd′=mq，
It can be obtained by folding it twice, OE=ON=OF=OM,
∴ob′=op=0d′=oq，
The quadrilateral B'PD'Q is a rectangle,
From folding, MN⊥EF, at point O,
∴PQ⊥B'D' is at 0 o'clock,
A quadrilateral is a square,