To get a negative answer for Q2, let $M$ be $L$, and consider the derived model at $\aleph_{\omega}$. For any set $A$ of ordinals, the Levy collapse $\mathrm{Levy}(\omega, A)$ is the partial order of containment on the set of finite partial functions $p \colon A \times \omega \to \bigcup A$ with $p(\alpha, i) \in \alpha$ for all $(\alpha, i)$ in the domain of $p$. Let $H$ be an $L$-generic filter for $\mathrm{Levy}(\omega, \aleph_{\omega})$, and let $G$ (which we would normally call $\mathbb{R}^{*}$) be $$\bigcup_{\gamma < \aleph_{\omega}} (\mathbb{R} \cap V[H \cap \mathrm{Levy}(\omega, \gamma)]).$$ Since $\mathrm{Levy}(\omega, \aleph_{\omega})$ is homogeneous, $\mathrm{HOD}^{L[H]} = L$. We want to see that $\aleph_{\omega}^{L} = \omega_{1}^{L(G)}$, as this will show that $\mathrm{DC}_{\mathbb{R}}$ fails in $L(G)$, so $\mathrm{DC}$ does as well. As forcing with $\mathrm{Levy}(\omega, \gamma)$ makes all ordinals less than $\gamma$ countable, $\aleph_{\omega}^{L} \leq \omega_{1}^{L(G)}$, so we want to see that there is no surjection from $\omega$ onto $\aleph_{\omega}^{L}$ in $L(G)$. Perhaps the easiest way to see this is to note that $L(G)$ is an inner model of $\mathrm{HOD}^{L[H]}_{G}$. Every real in $\mathrm{HOD}^{L[H]}_{G}$ is (by the definition of $\mathrm{HOD}^{L[H]}_{G}$) ordinal definable in $L[H]$ from a finite $g \subseteq G$, and any such $g$ is an element of $V[H \cap \mathrm{Levy}(\omega, \gamma)]$ for some $\gamma < \aleph_{\omega}^{L}$. Fix such a $\gamma$. By the homogeneity of $\mathrm{Levy}(\omega, \aleph_{\omega} \setminus \gamma)$ (and the fact that $\mathrm{Levy}(\omega, \aleph_{\omega})$ is isomorphic to $\mathrm{Levy}(\omega, \gamma) \times \mathrm{Levy}(\omega, \aleph_{\omega} \setminus \gamma)$), every real which is ordinal definable from $g$ in $L[H]$ must already be in $V[H \cap \mathrm{Levy}(\omega, \gamma)]$, in which $\aleph_{\omega}^{L}$ is uncountable.
For the last question in the comments, suppose that $X \in L(X)$ and that $L(X) \models \mathrm{DC}_{X}$. Suppose that $T$ is a tree of height $\omega$ without terminal nodes. There exists an ordinal $\gamma$ such that each element of $T \cup \{T\}$ is ordinal definable in $L_{\gamma}(X)$ from a finite subset of $X$. Fixing a list $\langle \phi_{i} : i < \omega \rangle$ of the first-order formulae, let $Y$ be the set of $(i, a, b) \in \omega \times X^{<\omega} \times \gamma^{<\omega}$ such that $\{ x \in L_{\gamma}(X) : L_{\gamma}(X) \models \phi_{i}(x,a,b)\}$ is an element of $T$. Let $T'$ be the tree order on $Y$ induced by $T$. An infinite branch through $T'$ then induces one for $T$.
It suffices then to show that for any set $X$, and any ordinal $\delta$, $\mathrm{DC}_{X}$ implies $\mathrm{DC}_{X \times \gamma}$. Fix a tree $T$ on $X \times \gamma$ without terminal nodes, and remove from $T$ each successor node $(\bar{x}^{\frown}\langle x_{n} \rangle, \bar{\alpha}^{\frown}\langle \alpha_{n} \rangle)$ for which there is a $\beta < \alpha$ with $(\bar{x}^{\frown}\langle x_{n} \rangle, \bar{\alpha}^{\frown}\langle \beta \rangle)$ in $T$. Call this tree $T'$, and let $T''$ be the tree formed from $T'$ be removing the second coordinate from each node. Then $T''$ has no terminal nodes, and an infinite branch through $T''$ induces infinite branches through $T'$ and $T$.