I'll give this a try. I certainly have an answer to Q2 and depending on what you mean exactly by boundary effects also an answer to Q1.

Q2. Yes. Take $\mathcal{C}$ as the unit disk. Player $A$ begins by choosing the origin. Then $B$ chooses some point $(b_1,b_2)$. Because of the rotational symmetry, we can assume $b_1=0$, $b_2>0$. Now one sees that $A$ controls the entire semidisk below the $x$-axis as well as a small strip up to the height $b_2/2$ of the upper semidisk. So $A$'s area is bigger.

Q1. No. Consider the same situation as above. We see that minimizing $b_2$ maximizes $B$'s area, so greedy play dictates that $b_2=\epsilon$. If your statement were true, then $A$ would continue by choosing $(0,2\epsilon)$, then $B$ chooses $(0,-\epsilon)$ etc.

Now suppose this has gone on for a while and the region of the disk between $y=-3/4$ and $y=3/4$ has been cut up into strips belonging to $A$ and $B$ respectively. (This certainly constitutes an "accumulation of $\epsilon$s". But we're still far away from the boundary if $\epsilon$ is small! So I'm not sure if this belongs already to what you mean by "boundary effects".)

Player $A$'s last move was to place a point at $(0,-3/4)$ (for simplicity $\epsilon=3/8n$ for some big $n$). If $B$ stays on the line and chooses $(0,-3/4-\epsilon)$ or $(0,3/4+\epsilon)$, it can gain at most around $0.1133...$ area.

But if $B$ chooses $(2/3,0)$, it will control at least the circle of radius 1/3 around this point. This circle has been cut up completely into strips and $A$ controls about half of it. So $B$ can gain about $\pi/18 \approx 0.1745...$ area.

I'll give this a try. I certainly have an answer to Q2 and depending on what you mean exactly by boundary effects also an answer to Q1.

Q2. Yes. Take $\mathcal{C}$ as the unit disk. Player $A$ begins by choosing the origin. Then $B$ chooses some point $(b_1,b_2)$. Because of the rotational symmetry, we can assume $b_1=0$, $b_2>0$. Now one sees that $A$ controls the entire semidisk below the $x$-axis as well as a small strip up to the height $b_2/2$ of the upper semidisk. So $A$'s area is bigger.

Q1. No. Consider the same situation as above. We see that minimizing $b_2$ maximizes $B$'s area, so greedy play dictates that $b_2=\epsilon$. If your statement were true, then $A$ would continue by choosing $(0,2\epsilon)$, then $B$ chooses $(0,-\epsilon)$ etc.

Now suppose this has gone on for a while and the region of the disk between $y=-3/4$ and $y=3/4$ has been cut up into strips belonging to $A$ and $B$ respectively. (This certainly constitutes an "accumulation of $\epsilon$s". But we're still far away from the boundary if $\epsilon$ is small! So I'm not sure if this belongs already to what you mean by "boundary effects".)

Player $A$'s last move was to place a point at $(0,-3/4)$ (for simplicity $\epsilon=3/8n$ for some big $n$). If $B$ stays on the line and chooses $(0,-3/4-\epsilon)$ or $(0,3/4+\epsilon)$, she can gain at most around $0.1133...$ area.

But if $B$ chooses $(2/3,0)$, she will control at least the circle of radius 1/3 around this point. This circle has been cut up completely into strips and $A$ controls about half of it. So $B$ can gain about $\pi/18 \approx 0.1745...$ area.