Answer: f2 - f1 = 132.52Hz.
Explanation: Since both tube are open at both ends, the length of air in the tube ( distance between antinode and antinode) is related to wavelength with the formulae below.
L = λ/2
considering the first tube, L1 = 12cm= 0.12m
λ1 = L1/2 = 0.12*2
λ1 = 0.24m
But v = f1 * λ1
350 = f1 * 0.24
f1 = 350/ 0.24
f1 = 1458.33Hz
Considering the second tube L2 = 11cm = 0.11m
L2 = λ2/2 = λ2 = 2* L2
λ2 = 2 * 0.11
λ2 = 0.22m
But v = fλ
350 = f2 * 0.22
f2 = 350/0.22
f2 = 1590.91
Frequency difference = f2 - f1 = 1590.91 - 1458.33
f2 - f1 = 132.52Hz