Using symmetry and Group theory argumens to explain $rm Fe^{2+}$ in a tetrahedral $rm Td$ crystal field

I am trying to figure out how to explain $1s rightarrow 3d$ spectroscopic transitions for $Fe^{2+}$ in $T_d$ symmetry. These transitions make up the pre edge region in K edge X-ray absorption spectroscopy (XAS). My goal is to rely on groupy theory and symmetry arguments as much as possible. Still, I am interested in the problem from a chemist perspective. I will first discuss what I already know and subsequently formulate my question.

$Fe^{2+}$ is a $d^6$ ion with the ground state Russel-Saunders term $^5D$. In the event of a $1s rightarrow 3d$ transition we find the final state configuration to be $1s^13d^7$. To find the corresponding terms we couple the core hole $s^1$ and the $d^7$ terms $^2S otimes {^4F, ^4P, ^2H, ^2G, ^2F, ^2P, ^2D(2)} = {^{5,3}F, ^{5,3}P, ^{1,3}H, ^{1,3}G, ^{1,3}F, ^{1,3}P, ^{1,3}D(2)}$. From the $^5D$ ground state only the quinted $F$ and quinted $P$ terns can be reached following the quadruople selection rule $Delta S = 0$. For the isolated ion (without spin orbit coupling) we thus have two transitions: $^5D rightarrow ^5F$ and $^5D rightarrow ^5P$. So far so good. Now we can include the $T_d$ crystal field.

In a $T_d$ crystal field the $5D$ terms splits into $^5E_u$ and $^5T_{2u}$. Of these, the $^5E_u$ represents the lowest energy $(e_2)^3(t_2)^3$ configuration. Our initial state irreducible representations $Gamma_i$ thus is $Gamma_i = ^5E$.
$$Gamma_i = ^5E$$
The final state irreducible representations $Gamma_f$ are obtained by branching the $^5P$ and $^5F$ terms from $O_3$ to $O_h$:
$$^5F rightarrow ^5A_{2u} oplus ^5T_{2u} oplus ^5T_{1u} $$
$$^5P rightarrow ^5T_{1u} $$
$$Gamma_f = ^5A_{2u} oplus ^5T_{2u}oplus ^5T_{1u}(F) oplus ^5T_{1u}(P)$$

The quadruople transition operator in $T_d$ symmetry is given by $Gamma_{hat{T}} = T_2 oplus E$. Now I understand that a transition is possible if the matrix element $<f|hat{T}|i>$ is non zero. To see which final states irreducible representations $Gamma_f$ are accesible trough the transtion operator $hat{T}$ we can take the direct product $Gamma_i otimes Gamma_{hat{T}}$. This will give:
$$^5E_g otimes (T_2 oplus E) = ^5E_g otimes T_2 oplus ^5E_g otimes E = ^5A_{1u} oplus ^5A_{2u} oplus ^5E_u oplus ^5T_{1u} oplus ^5T_{2u}$$

Please correct me if I already made some mistakes. From here I am not so sure how to continue. All the final state irreps $Gamma_f$ are contained in the direct product $Gamma_i otimes Gamma_f$.

1. Can i conclude that from the $^5E_g$ ground state all of the final state irreps $Gamma_f$ can and will be reached?
2. If the only possible final states are $(e_2)^4(t_2)^3$ and $(e_2)^3(t_2)^4$ which $Gamma_f$ irreps correspond to these configurations?
3. It is well known that the K pre edge gains intensity for non-centrosymmetric ions with respect to centrosymmetric ions. This is because the $4p$ orbitals can mix with the $d_{xy}$, $d_{xz}$ and $d_{yz}$ orbitals. How does this fit the above storyline?