(1)由题设知Sn=2-2an+1,从百推导出2an+1=an,即,再由a1=1,能求出a2,a3.
(2)由2an+1=an,知=,再由a1=1,能求出an.
【解析】
(1)数列{an}的前n项和为Sn,a1=1,
且对任意正整数n,点(an+1,Sn)在直线2x+y-2=0上,
∴Sn=2-2an+1,
∴an=Sn-Sn-1=(2-2an+1)-(2-2an)=2an-2an+1,
∴2an+1=an,
∴,
∴=,
=.
(2)∵2an+1=an,∴=,
∴a1=1,∴an=()n-1.