Given equations:
A + 1/B = 3 ...(i)
B + 1/A = 5 ...(ii)

From equation (i):
A = 3 - 1/B ...(iii)

Substitute equation (iii) into equation (ii):
3 - 1/B + 1/A = 5
3 - 1/B + 1/(3 - 1/B) = 5
3 - 1/B + B/(3B - 1) = 5

Multiply through by B(3B - 1):
3B(3B - 1) - B + B^2 = 5B(3B - 1)
9B^2 - 3B - B + B^2 = 15B^2 - 5B
10B^2 - 8B = 15B^2 - 5B

Solve for B:
10B^2 - 8B = 15B^2 - 5B
-5B^2 - 3B = 0
B(-5B - 3) = 0
B = 0 or B = -3/5

Since B cannot be 0, B = -3/5

Now substitute B = -3/5 into equation (iii):
A = 3 - 1/(-3/5)
A = 3 - 5/(-3)
A = 3 + 5/3
A = 14/3

Now, (A + B)/(A - B) = (14/3 - 3/5)/(14/3 + 3/5) = (67/15)/(73/15) = 67/73

Therefore, (a+b)/(a-b) = 67/73.