(1)由题知(p-1)Sn=p²-an ①,
则(p-1)Sn-1=p²-an-1 ②,
①-②,得(p-1)an=an-1-an,即p*an=an-1,an/an-1=1/p,
当n=1时,(p-1)S1=p²-a1,即(p-1)a1=p²-a1,故a1=p,
因为a3=1/3=a1*(1/p)²=1/p,所以p=3=a1,1/p=1/3,
所以{an}是以3为首项,1/3为公比的等比数列,其通项公式为an=(1/3)^(n-2)=3^(2-n).
(2)bn=1/(2-log3an)=1/[2-(2-n)]=1/n,
//如果你说的数列﹛bnbn＋2﹜是{bn*b(n+2)}的话
bn*b(n+2)=1/n(n+2)=1/2[1/n-1/(n+2)],
Tn=b1*b3+b2*b4+…+bn*b(n+2)
=1/2[1-1/3+1/2-1/4+1/3-1/5+…+1/(n-1)-1/(n+1)+1/n-1/(n-2)]
=1/2[1-1/(n-2)],
故Tn的极限值为1/2,则当m²-m＋3/4≥1/2时,满足对任意正整数n,都有Tn＜m²-m＋34成立,
m²-m＋3/4≥1/2m²-m＋1/4≥0
(m-1/2)²≥0
因为(m-1/2)²≥0恒成立,所以m的取值范围为R