(1)依题意可求得等比数列{an}的公比q,利用其通项公式即可求得an;
(2)由(1)知an=,于是可求得bn=-n,易证数列{bn}是以为首项,-1为公差的等差数列,从而可求其前n项和Sn.
【解析】
(1)∵等比数列{an}的公比q>0,a1是3a2与2a3的等差中项,
∴3a1q+2a1q2=2a1,又a1=,
∴q=或q=-2(舍去),
∴an=;
(2)∵bn=+log2an=+log2=-n,
∴bn+1=-(n+1),
∴bn+1-bn=-1,又b1=,
∴数列{bn}是以为首项,-1为公差的等差数列,
∴Sn=b1+b2+…+bn
=n+×(-1)
=-n2+10n.
=-(n-10)2+50,
∴当n=10时,S10取得最大值50.