( 1)

Let {an} tolerance be d, then d≠0. Let the common ratio of {bn} be q.

a2=b2，a 1+d=b 1q

If a 1=b 1= 1, then d+1= q.

d=q- 1

D≠0, then q≠ 1.

a8=b3，a 1+7d=b 1q？

If a 1=b 1= 1, 7d+ 1=q?

d=(q？ - 1)/7

q- 1=(q？ - 1)/7

Tidy up and get a Q? -7q+6=0

(q- 1)(q-6)=0

Q= 1 (truncated) or Q=6.

d=q- 1=6- 1=5

The tolerance of sequence {an} is 5, and the common ratio of sequence {bn} is 6.

(2)

an = a 1+(n- 1)d = 1+5(n- 1)= 5n-4

bn=b 1q？ = 1 6? =6?

The general formula of sequence {an} is an=5n-4, and the general formula of sequence {bn} is bn=6? .

(3)

sn =(a 1+an)n/2 =( 1+5n-4)n/2 = n(5n-3)/2

Tn=b 1(q？ - 1)/(q- 1)= 1(6？ - 1)/(6- 1)=(6? - 1)/5

The sum of the first n terms of sequence {an} is n(5n-3)/2, and the sum of the first n terms of sequence {bn} is (6? - 1)/5。