1) Like we can see in the picture, the splitting due to the applied field will cause 4 new levels. Only transitions which obey the selection rules can happen (|\Delta m_J| = 1), and since the splitting between neighbouring levels are the same, only one frequency will be absorbed. This would be the one with energy g_J*µ_B*\|\Delta m_J|*B = 0.8*5.788(eV/T)*1*2T = 92,61 µeV.
2) The levels are again splitted symmetrically around their original level, but now with an energy difference of A*m_I*m_J. The extra selection rule |\Delta m_I|=0 now also has to be satisfied. We will calculate following transitions (notation = (m_J, m_I):
a) (-3/2, 1/2) to (-1/2, 1/2)
b) (-3/2, -1/2) to (-1/2, -1/2)
c) (-1/2, 1/2) to (1/2, 1/2)
d) (-1/2, -1/2) to (1/2, -1/2)
Because of the symmetrical splitting, we can already conclude that c and d will be the same.
The energy levels have an energy E(m_I m_J) = -g_J*µ_B*m_J*B + A*m_I*m_J with A=g*\mu_N*B_hf/J. From this we can calculate the differences with ease:
\DeltaE = -g_J*µ_B*B*(m_J-m_J’) + A*m_I*(m_J-m_J’)
a) 46,25 µeV
b) 46,35 µeV
c=d) 92,56 µeV