(1) Substitute the coordinates of two points into the function expression to get two equations: ab? = 1/2，ab？ = 1;

Solution: a= 1/8, b = 2；;

Analytical formula of function y=f(x): f (x) = (2 x)/8;

(2) an = log {2} f (n) = log {2} [(2n)/8] = n-3, so {an} is arithmetic progression, the first term is a 1=-2, and the tolerance is d =1;

∴sn = a 1+(n- 1)* d =-2+(n- 1)* 1 = n+ 1；

An*Sn=(n-3)*(n+ 1)≤0, and the solution is n ≤ 3;

(3)an*f(n)=(n-3)*(2^n)/8=(n-3)*2^(n-3)； tn=∑(n-3)*2^(n-3),tn*2=∑(n-3)*2^(n-2)；

2tn-tn=∑(n-3)*2^(n-2)-∑(n-3)*2^(n-3)；

=2*2^(-2)-2^(- 1)-2^0-2^ 1-2^2-2^3-……-2^(n-3)+(n-3)*2^(n-2)；

∴tn=(n-3)*2^(n-2)+2*2^(-2)-[2^(- 1)-2^(n-2)]/( 1-2)=(n-3)*2^(n-2)+( 1/2)-[2^(n-2)-( 1/2)]

=(n-4)*2^(n-2)+ 1；