How does a d4 high-spin complex have a spin-allowed transition without violating the selection rule for total angular momentum?
The photon has one unit of angular momentum and as angular momentum is conserved this must be taken up by the molecule; in terms of symmetry species this means that the product of the symmetry species of the two states involved and of the transition dipole must be worked out. The electron spin in not affected by the photon angular momentum so the spin selection rule applies $\Delta S=0$ unless there is also some spin - orbit coupling then otherwise spin-forbidden transitions can occur but with generally low probability compared to a spin allowed one.

To work out the symmetry species involved in a transition the value of the integral 

$$\int\psi_i\mu\psi_fd\tau$$ 

between initial $i$ and final state $f$ has to be evaluated.; $\mu$ is the transition dipole. This integral is very hard to evaluate but fortunately symmetry can help.

Using symmetry the integral can be determined to be zero (transition forbidden) or not zero. The equation to use is 

$$\Gamma\psi_i \Gamma\mu \Gamma\psi_f = \;'A_1\,'$$

where $\Gamma$ means the symmetry species, $A,\; E$ etc and $'A_1\,'$ means the product must contain the totally symmetric representation in the point group used. This is the label given to the top row of characters in the table ($A_1$ or $A_{1g}$). 

The transition dipole symmetry is found in the column of the point group just to the right of the characters and are the symmetry species with x, y, and z. In the $T_d$ and $O$ types of point groups the transition dipoles are either the $T_2$ or $T_1$ symmetry species respectively.

Thus in your $A_1\to T_1$ example the symmetry product is 

$$A_1 \otimes T_1 \otimes T_1 = A_1$$

in an $O$ point group. The symbol $\otimes$ means multiply characters in each symmetry species column by column.  Hence you can work out the product  by multiplying the characters together or more easily by looking up the values in direct product tables. 

In this case this is easy to do as anything multiplied by $A_1$ is unchanged (as its totally symmetric) and any symmetry species multiplied by itself always contains the totally symmetric representation $A_1$. So the transition 
$A_1\to T_1$ is allowed in octadedral point groups.

In the tetrahedral point group $T_d$ where the transition dipoles each transform as $T_2$ the product is 

$$A_1 \otimes T_2 \otimes T_1 =  T_2 \otimes T_1$$

and now the characters have to be multiplied and the the irreducible representation found (see link to answer below). Instead looking up the product in direct product tables gives

$$T_2 \otimes T_1 = A_2+E+T_1+T_2$$

and in this case as the totally symmetric representation $A_1$ is absent the transition would not be allowed. This means that any transition will be weak, i.e. has a very low but finite extinction coefficient. There are effects in molecules that 'break' the symmetry, such as vibrations distorting the molecules 'ideal' symmetry and these can make forbidden transitions slightly allowed.

This answer gives some details on point groups.
Understanding group theory easily and quickly