(2)过B作BF⊥AD,过C作CE⊥AD,
∵点A、B的横坐标分别是a和a+2,
∴可得,A(a,3a 2 -7),B(a+2,3a 2 -4),
C(a+2,12 a+2 ),D(a,12 a ),
∵AB=CD,
∴在Rt△CDE与Rt△ABF中,
由勾股定理得:CD2=DE2+EC2=22+(12 a+2 -12 a )2,
AB2=AF2+BF2=22+32,
∵等腰梯形ABCD,
∴AB=CD,即22+32=22+(12 a+2 -12 a )2,
即12 a+2 -12 a =±3,
①由12 a+2 -12 a =3,化简得a2+2a+8=0,方程无实数根,
②由12 a+2 -12 a =-3,化简得a2+2a-8=0,
∴a=-4,a=2,
经检验,a=-4,a=2均为所求的值.