(1)∵四边形ABCD是矩形,AB=4,BC=3,
∴AC=
AB 2 + BC 2 =
4 2 + 3 2 =5,
∵△B′CE由△BCE翻折而成,
∴B′C=BC=3,BE=B′E,
∴AB′=AC-B′C=5-3=2;
设AE=x,则BE=B′E=4-x,
在Rt△AB′E中,AE 2=AB′ 2+B′E 2,即x 2=2 2+(4-x) 2,解得x=
5
2 ,即AE=
5
2 ;
(2)∵由(1)知,AE=
5
2 ,
∴S △AEC=
1
2 AE•BC=
1
2 ×
5
2 ×3=
15
4 .