(1)∵等差数列{an}的前n项和为Sn,且满足条件S8=36,a3=3,
∴
8a1+
8×7
2d=36
a1+2d=3,
解得a1=1,d=1,∴an=n.
(2)∵bn=[1
an+
1
an+1+…+
1
a2n=
1/n+
1
n+1+…+
1
2n],
∴bn+1-bn=[1/2n+2+
1
2n+1−
1
n]
=([1/2n+2−
1
2n])+([1/2n+1−
1
2n])<0,
∴{bn}为递减数列,∴(bn)max=b1=1+[1/2]=[3/2],
∵log2(
1
4x2+x)−bn>0恒成立,
∴log2(
1
4x2+x)>bmax=
3
2,
∴
1
4x2+x>2
3
2=2