For (1), the point is that $\text{cof}(\omega)\cap\lambda$ is the set of ordinals below $\lambda$ that have cofinality $\omega$ in $V$, but this is not necessarily the set of ordinals below $\lambda$ that have cofinality $\omega$ in $\text{HOD}$, since having cofinality $\omega$ is not necessarily absolute between $V$ and $\text{HOD}$. So this is a definable set of ordinals, and this puts it into $\text{HOD}$, but there is in general no reason to think that it is the same as $\text{cof}(\omega)^{\text{HOD}}\cap\lambda$, that is, the set of ordinals below $\lambda$ that have cofinality $\omega$ in $\text{HOD}$. It is just a set of ordinals, and it includes every ordinal below $\lambda$ in $\text{HOD}$ that has cofinality $\omega$ in $\text{HOD}$, but it may also include other ordinals, which have cofinality $\omega$ in $V$ but not in $\text{HOD}$.
For (2), being club is absolute between $\text{HOD}$ and $V$, since for $C\subset\kappa$ to be club means that it is closed and unbounded in $\kappa$, and if you think about it, you will realize that $V$ and $\text{HOD}$ cannot disagree about whether a set is closed or whether it is unbounded in $\kappa$. Meanwhile, being stationary is not necessarily absolute, since a set is stationary when it has nonempty intersection with every club, but $\text{HOD}$ may simply have fewer club sets, and so it is easier for a set to be stationary in $\text{HOD}$ than in $V$. This can definitely happen, if you force over $L$, say to destroy the stationarity of the $L$-least stationary co-stationary subset $S\subset\omega_1$. This forcing is homogeneous, and so the $\text{HOD}$ of the extension $L[G]$ is just $L$ again, where $S$ is stationary, even though it isn't stationary in $L[G]$. Meanwhile, being stationary is downward absolute.
For (3), what Woodin means is just that the set $\{S\in\text{HOD}\mid S\text{ is stationary in }V\}$ is a set in $\text{HOD}$, and so $\text{HOD}$ has access to this set. So $\text{HOD}$ can know which of its stationary sets are actually stationary in $V$, since the set of those sets is definable in $V$ and hence an element of $\text{HOD}$.