1a) For 111Cd, we have for the ground state: I = 1/2 and mu = -0.5941 nm. Using the formula \mu = (g\mu_n)/\hbar I, we find g = -2*0.5940 nm*\hbar/\mu_N. For the 245 keV level: I = 5/2 and \mu = -0.766 nm and so we find g = -(2/5)*0.766*\hbar/\mu_N.
1b) Free electron: the spin is 1/2, and mu = 1 \mu_B, so we find g = 2\hbar*(\mu_N/\mu_B)
1c) Free neutron: we find \mu = (-3.826/2) \mu_N/\hbar.