(1)∵在Rt△ABC中,∠ACB=90°,∴sinA= BC AB = 4 5 ,设BC=4k,则AB=5k,∴AC= AB2?BC2 =3k,∵AC=6,∴3k=6,k=2,∴AB=10,∵D是边AB的中点,∴CD= 1 2 AB=5;(2)过C点作CF⊥AB于F.CF=AC?BC÷AB=4.8,cos∠DCF= CF CD = 4.8 5 = 24 25 .∵∠DCM=∠DBE,∴cos∠DBE= 24 25 .