如图所示:(1) 设AM=x>0,AB=y>0,则DM^=1+x^,CM^=1+(y-x)^,当CM⊥DM时,1+x^+1+(y-x)^=y^===>x^-yx+1=0,∵ M点唯一,∴ 判别式△=0,得y=2,x=1.∴当AB=2且M是AB的中点时,CM⊥DM.D'D⊥面ABCD,DM是D'M在面ABCD的射影,由三垂线逆定理,CM⊥D'M 又C'C⊥CM,∴ CM是异面直线C'C与D'M的距离=√2.(2) △D'DC的面积=1,D'M=√3,CM=√2,∴△D'MC的面积=√6/2,△D'MC在面D'DCC'的射影是△D'DC.二面角M-D'C-D的平面角为θ,由面积射影定理,得cosθ=△D'DC/△D'C=√6/3,∴ 二面角M-D'C-D的大小 是arccos(√6/3).