(1)由已知∠AEB=∠BFC=90°,AB=BC,
又∵∠ABE+∠FBC=∠BCF+∠FBC,
∴∠ABE=∠BCF,
∵在△ABE和△BCF中,
AB=BC
∠ABE=∠BCF
∠AEB=∠BFC,
∴△ABE≌△BCF(AAS),
∴AE=BF,
∴AE2+CF2=BF2+CF2=BC2=16为常数;
(2)设AP=x,则PD=4-x,
由已知∠DPM=∠PAE=∠ABP,
∴△PDM∽△BAP,
∴[DM/AP]=[PD/BA],
即[DM/4?x]=[x/4],
∴DM=
x(4?x)
4=x-[1/4]x2,
当x=2时,即点P是AD的中点时,DM有最大值为1.