Crystal field theory

$Co^{3+}$: $[Ar]3d^6$
$CN^-$ is a strong field ligand, so $\Delta_o$ is large and the electrons pair up in the $t_{2g}$ level.
[add image: $3d^6$ splitting into $t_{2g}^6 e_g^0$ under a strong field]
Configuration $t_{2g}^6e_g^0$: no unpaired electrons, low-spin complex.

$Fe^{3+}$ is $d^5$ and $H_2O$ is a weak field ligand, so the complex is high spin: $t_{2g}^3e_g^2$.
CFSE $= 3(-0.4\Delta_o) + 2(0.6\Delta_o) = 0$

In a low spin $d^6$ octahedral complex all six electrons are paired in the $t_{2g}$ orbitals: $t_{2g}^6e_g^0$.
Each $t_{2g}$ electron is stabilised by $-\frac{2}{5}\Delta_0$ ($= -0.4\Delta_0$) and each $e_g$ electron is destabilised by $+\frac{3}{5}\Delta_0$.
Energy from splitting $= 6\times\left(-\frac{2}{5}\Delta_0\right) + 0 = -\frac{12}{5}\Delta_0$
There are three electron pairs, so the pairing energy is 3P.
Total energy $= -\frac{12}{5}\Delta_0 + 3P$

In an octahedral field the d orbitals split into $t_{2g}$ (each electron stabilised by $-0.4\Delta_0$) and $e_g$ (each electron destabilised by $+0.6\Delta_0$).
High spin $d^4$ means the fourth electron goes to $e_g$ instead of pairing: configuration $t_{2g}^3e_g^1$.
CFSE = $3\times(-0.4\Delta_0) + 1\times(+0.6\Delta_0)$
CFSE = $-1.2\Delta_0 + 0.6\Delta_0$
CFSE = $-0.6\Delta_0$
(No pairing energy term appears, because no electrons are paired.)