∠∠ 1 = 180-∠AMB-∠AMN，∠2= 180 -∠AMB-∠B，∠AMN=∠B=60，
∴∠ 1=∠2.
And ∵CN divides ∠ACP, ∠ 4 = 12 ∠ ACP = 60,
∴∠MCN=∠3+∠4= 120。 ①
And ∵BA=BC, EA=MC,
Ba-ea = BC-MC, which means be = BM.
△ BEM is an equilateral triangle.
∴∠6=60 .
∴∠5= 180 -∠6= 120 .②
∴ MCN =∠ 5 comes from ① ②.
At △AEM and △MCN,
∠ 1=∠2AE=CM∠MCN=∠5，
∴△AEM≌△MCN(ASA).
∴AM=MN.
So the answer is: ∠1= ∠ 2; AE = CM∠MCN =∠5；
(2) Solution: Conclusion A1m1= m1n1is still valid for the following reasons:
Intercept A 1E=C 1M 1 on A 1B 1 and connect EM 1.
The quadrangle a1b1c1d1is a square.
∴a 1b 1=b 1c 1,∠b 1=90，
∴EB 1=M 1B 1, that is, △EB 1M 1 is an isosceles right triangle.
∴∠B 1EM 1=45，
∫c 1n 1 is the bisector of∠ d1c1p1.
∴∠a 1em 1=∠n 1c 1p 1= 135，
≈b 1a 1m 1+≈a 1m 1b 1 = 90，≈a 1m 1b 1+≈n 1m 1c 1 = 90，
∴∠b 1a 1m 1=∠n 1m 1c 1，
In △A 1EM 1 and △ m1n1,
∠ea 1m 1 =∠c 1m 1n 1E = m 1c 1∠a 1em 1 =∠m 1c 1n 1，
∴△a 1em 1≌△m 1n 1c 1(asa)，
∴a 1m 1=m 1n 1.