解答:(1)证明:∵四边形ABCD是正方形,∴∠A=∠B=90°,AB=BC=AD,∵E是AB中点,FC=3BF,∴AD:BE=AE:BF=2,∴△BEF∽△ADE;(2)△ADE∽△EDF.证明:∵△BEF∽△ADE,∴∠AED=∠BFE,AD:BE=DE:EF,∵AE=BE,∴AD:DE=AE:EF,∵∠BEF+∠BFE=90°,∴∠AED+∠BEF=90°,∴∠DEF=∠A=90°,∴△ADE∽△EDF.