(1) take the midpoint d of A 1B 1 and connect BD, DE, which is the center line of triangle a1b1,DE is parallel to BF, and quadrilateral BFED is a parallelogram, so BD is parallel to EF. Therefore, EF∑ plane aa1b1b.

(2) According to the first question, the angle formed by EF and plane ABC is the angle formed by BD and AB, which is equal to the angle BDB 1. In the right triangle BDB 1, BB 1=3 and B 1D=√3, the angle is 60.