If $\kappa$ is an inaccessible cardinal and $G \subset \operatorname{Col}(\omega,\mathord{<}\kappa)$ is a $V$-generic filter, then in $V[G]$ the Chang model $L(\text{Ord}^\omega)$ satisfies "every set of reals has the classical regularity properties" by Solovay's theorem. (By the classical regularity properties I mean Lebesgue measurability, the Baire property, and the perfect set property.)

If instead $\kappa$ is a limit of Woodin cardinals (not necessarily inaccessible, possibly singular,) $G \subset \operatorname{Col}(\omega,\mathord{<}\kappa)$ is a $V$-generic filter, and we define $\mathbb{R}^*_G = \bigcup_{\alpha < \kappa} \mathbb{R}^{V[G\restriction \alpha]}$, then the model $L(\mathbb{R}^*_G)$ satisfies the Axiom of Determinacy (and hence also the weaker statement that every set of reals has the classical regularity properties) by a theorem of Woodin.

Now let's assume that $\kappa$ is inaccessible *and* is a limit of Woodin cardinals. Again let $G \subset \operatorname{Col}(\omega,\mathord{<}\kappa)$ be a $V$-generic filter. Because $\kappa$ is inaccessible we have $\mathbb{R}^*_G = \mathbb{R}^{V[G]}$. In this case it is possible to obtain a mutual strengthening of the two results mentioned above:

In $V[G]$, does the Chang model $L(\text{Ord}^\omega)$ satisfy $\mathsf{AD}$?

(I am aware that Woodin has proved that the much stronger hypothesis of the existence of a proper class of Woodin limits of Woodin cardinals implies that the Chang model of $V$ satisfies $\mathsf{AD}$.)