Consistency strength of strongly compact cardinal

Question

1. Where can I find a proof that strongly compact cardinal has higher consistency strength than Woodin cardinal, or even just strong? Recall that a strongly compact cardinal itself may not be strong, since consistently the least measurable cardinal is strongly compact. I checked Jech and Kanamori and they merely mention that strongly compact implies inner model with many measurables.

2. Does existence of a strongly compact cardinal imply $\mathrm{AD}^{L(\mathbb{R})}$?

Answer 1

Steel [1] showed that if $\square_\kappa$ fails for some singular strong limit $\kappa$, then $\text{AD}$ holds in ${L(\mathbb R)}$. Since Solovay showed a strongly compact cardinal implies the failure of $\square_\kappa$ for all sufficiently large $\kappa$, we have that a strongly compact cardinal implies $\text{AD}^{L(\mathbb R)}$. If you just want the consistency of a Woodin cardinal from a strongly compact, you get a much easier proof by combining Corollary 2.5.21 in Larson's Stationary Tower with Theorem 7.1 in Steel's Core Model Iterability Problem. Larson's Corollary 2.5.21 (due to Woodin) says that if $\kappa$ is strongly compact, there are many $\gamma < \kappa$ such that if $G$ is $V$-generic for the full stationary tower $\mathbb P_{<\gamma}$, then in $V[G]$ there is an elementary embedding $j : V \to M$ such that $\text{crit}(j) < \gamma$, $j(\text{crit}(j)) < \gamma$, and $M^{<\gamma}\cap V[G]\subseteq M$. Steel's Theorem 7.1 says that if a forcing of size less than a measurable $\kappa$ adds such an embedding, then $V_\kappa$ has an inner model with a Woodin cardinal.