1.a)
We know that µ = g * µ_N * I / hbar, so g = hbar * µ / µ_N * I.
For the ground state of 111Cd we have µ = -0.594 µ_N and I = 1/2 hbar, so g = -1.188.
For the state with energy 245 keV we have µ = -0.766 µ_N and I = 5/2, so g = -0.306.

1.b)
We now have an analogous formula for g: g = µ * hbar / µ_B * I. For the free electron: µ = 1 µ_B and I = 1/2 hbar, so g = 2.

1.c)
We can convert the above formula: µ = g * I * µ_N / hbar. With g = -3.826 and I = 1/2 hbar we find µ = -1.913 µ_N.