1. Because MA is equiangular to AB and AD, it can be proved that MA is projected on the plane ABCD to BD. Cos ∠ MAC = ∠ 2/4 can be calculated by the three perpendicular theorem (in fact, you can read a book and there should be a conclusion similar to "cos∠MAC=cos∠MAB×cos∠BAC"). In △MAC, the length of MC can be calculated by cosine theorem.

2, the second question, I really can't think of a good way. The following methods are for your reference: take the midpoint H of MA and connect HN and HB, then in △HNB, ∠HNB is the angle formed by non-planar straight line AC and BN or the remaining angle. ① at △ AHB, solve the length of HB; ②NH is half of AC, so solve; (3) BN is the most uncomfortable thing for me and provides the most stupid method (sorry! ! ! ): In △ABM, get BM, in △ACM, get CM. Extend bn twice to generate a parallelogram with 2 (BM 2+BC 2) = MC 2+(2bn) 2, calculate the length of BN, and then calculate the cosine of ∠HNB at △HNB.

I really can't think of any other good method for the second question. . . . . . . . .

In addition: because this problem lacks the foundation (vertical line) to establish a spatial coordinate system, I personally feel that it should not be measured.