(1) When the angle c is equal to 45 degrees, the triangle ABC is an isosceles triangle, and the endpoint n of the median vertical line MN coincides with a, and AM=MC. Because angle PNM+ angle MNC = 90, angle MNC+ angle MNC = 90, angle PNM= angle NCM, similarly, angle PMN= angle QMC. So the triangle PNM is equal to the triangle QCM, so PA=QC.

So PA+QN=QC+QN=QA=sqrt(2)/2*BC.

(2)AP+NQ & lt； BC 1/2 *

According to the problem, the angle CNM= angle PBM = 60°, because the angle QMN+ angle PMB = 90, the angle NMP+ angle PMB = 90, and the known angle QMN= angle PMB, so the triangle NQM is similar to the triangle BPM, and because in the triangle NCM, the angle C = 30, so NM.

(3)