I think the following may be relevant to your question:

First, several bullet points in your question are nontrivial results. However, there is another way to see $\mathsf{AD}_\mathbb{R}$ does not holds in $L(\mathbb{R})$.

Using a two step game, you can show $\mathsf{AD}_\mathbb{R}$ can prove uniformization for all sets of reals. Consider the relation $R(x,y)$ if and only if $y$ is not ordinal definable from $x$. Assuming that $L(\mathbb{R}) \models \mathsf{AD}_\mathbb{R}$, $R$ would have a uniformization. Every set of reals in $L(\mathbb{R})$ is ordinal definable from some real. Use this real to get a contradiction by diagonalization.

Also note that if you put the discrete topology on $\mathbb{R}$, a two point game is a closed game. Hence $\mathsf{ZF + AD}$ can not prove the Gale-Stewart theorem (closed determinacy) for games on $\mathbb{R}$. Of course, this is because there is no well-ordering of $\mathbb{R}$.

"The Extent of Scales in $L(\mathbb{R})$" Corollary 6 shows that $\mathsf{DET}_\mathbb{R}(\mathbf{\Pi}_1^1)$ can not hold in $L(\mathbb{R})$. Steel shows this by proving that $G^\mathbb{R}\Pi_1^1$ (the game quantifier class) is $\Sigma_1(L(\mathbb{R}, \{\mathbb{R}\})$. By $\mathsf{DET}_\mathbb{R}(\mathbf{\Pi}_1^1)$, $G^\mathbb{R} \Sigma_1^1$ is the dual of $G^\mathbb{R}\Pi_1^1$. Hence $G^\mathbb{R}\Sigma_1^1$ is $\Pi_1(L(\mathbb{R}, \{\mathbb{R}\})$. But you can directly show that $G^\mathbb{R}\Sigma_1^1$ is contained in $\Sigma_1(L(\mathbb{R}),\{\mathbb{R}\})$. Contradiction.

($\Sigma_1(L(\mathbb{R},\{\mathbb{R}\})$ is also $(\Sigma^2_1)^{L(\mathbb{R}}$. Using Wadges lemma and the not-ordinal-definable relation $R$ from above, Steels showed that if $\mathsf{AD}^{L(\mathbb{R})}$ holds, then a set has a scale in $L(\mathbb{R})$ if and only if it belongs to $(\mathbf{\Sigma}^2_1)^{L(\mathbb{R})}$. )

The same paper of Steel also notes that assuming $\mathsf{AD}$, $\mathsf{DET}_\mathbb{R}(\mathbf{\Pi}_1^1)$ is equivalent to the existence of $\mathbb{R}^\sharp$. (At least in the paper, Steel states it is open whether the equivalence can be proved in just $\mathsf{ZF} + \mathsf{DC}$.) Assuming $\mathsf{AD}$, $\mathsf{DET}_\mathbb{R}(\mathbf{\Pi}_1^1)$ has stronger consistency strength than $\mathsf{AD}$.