【解】：
S[n]=a[n+1]-3n-1；S[n-1]=a[n]-3(n-1)-1=a[n]-3n+2
a[n]=S[n]-S[n-1]=(a[n+1]-3n-1)-(a[n]-3n+2)=a[n+1]-a[n]-3
得：a[n+1]=2a[n]+3
即：(a[n+1]+3)=2×(a[n]+3)
所以{a[n]+3}为以2为公比的等比数列,且a[1]+3＝4；
则：a[n]+3=4*2^(n-1)=2^(n+1)
得：a[n]=2^(n+1)-3
得：b[n]=2^n/(2^(n+1)-3)(2^(n+2)-3)=2/(2^(n+1)-3)-1/(2^(n+2)-3)
b[1]=2/(2^2-3)-2/(2^3-3)
b[2]=2/(2^3-3)-2/(2^4-3)
b[3]=2/(2^4-3)-2/(2^5-3)
……
b[n-1]=2/(2^n-3)-2/(2^(n+1)-3)
b[n]=2/(2^(n+1)-3)-2/(2^(n+2)-3)
相加消项得：T[n]=2/(2^2-3)-2/(2^(n+2)-3)=2-2/(2^(n+2)-3)
|(T[n]-1)/2|1/2010.
显然题目这个有点问题,不知何意.