Horton equation:

f = f_c + (f_o - f_c) e^-kt

f - f_c = (f_o - f_c) e^-kt

3 - 1 = (f_o - 1) e^-k

2 - 1 = (f_o - 1) e^-2k

2 = (f_o - 1) e^-k

1 = (f_o - 1) e^-2k

Dividing these two equations:

2 = e^k

k = ln (2) = 0.6931 hr^-1    ANSWER.

Solving for f_o:

f_o = 2 e^k + 1

f_o = (2 × 2) + 1 = 5 mm/hr    ANSWER.

f_c = 1 mm/hr    ANSWER.

Verification:

f = 1 + (5 - 1) e^-0.6931 = 3 mm/hr   OK

f = 1 + (5 - 1) e^{-2 × 0.6931} = 2 mm/hr   OK