Understanding energy level splitting of an octahedral complex from spin-orbit coupling, 1st and 2nd order zeeman effects
To answer part b it is necessary to consider the symmetry properties in an octahedral field. The spin orbit Hamiltonian is $H_i=\zeta\sum_i\bar l_i\cdot \bar s_i$ where $\zeta$ is a constant characteristic of the radial part of the d orbitals. The orbital angular momentum, which may be written in the form (for the $x$ component) as $\displaystyle l_x =-i\hbar \left(y\frac{\partial }{\partial z}-z\frac{\partial}{\partial y}\right)$ (and similarly for $l_y$ and $l_x$ by rotating subscripts) transforms as does the $T_{1g}$ irreducible representation in an $O_h$ field. 

If a term has an orbital symmetry $X$ it will only be susceptible to spin orbit coupling if the symmetry product $X\cdot T_{1g}\cdot X$ contains the totally symmetric representation $A_{1g}$. In other notation $\langle i|H |j\rangle \ne 0$ or $\int \psi_i H \psi_f d\tau \ne 0$). Thus the $E_g$ levels are unaffected in point group $O_h$ because $E_g \cdot T_{1g}\cdot E_g$ does not contain $A_{1g}$ but the $T_{2g}$ interact ($T_{2g} \cdot T_{1g}\cdot T_{2g}$ does  contain $A_{1g}$).

Notes: It is assumed that the potential energy effect of the ligands is only small compared to that of the central field of the ion, thus the spin-orbit coupling is a small perturbation on the energies. 

The full derivation is complex and involves using double symmetry groups.

The free ions have SO couplings $^2\mathrm D_{3/2} =-3\zeta_d/2$ and $^2\mathrm D_{5/2} =\zeta_d$. (The formula you use if for the free ion).  In the octahedral field the calculation is rather complicated but the result is that $T_{2g}$ levels are split by $3\zeta/2$ in the ratio you show.  

In the remaining parts of your question similar symmetry arguments should indicate what levels are split, but only a detailed calculation will produce the energies that they are split by.