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As dke mentioned, I have a paper in which I construct various kinds of varieties $X_{/\mathbb{Q}}$ without abelian points (i.e., with $X(\mathbb{Q}^{\operatorname{ab}}) = \varnothing$). Here is a brief summary:

If $X$ admits a $2:1$ map to a variety $Y$ with infinitely many $\mathbb{Q}$-rational points, then $X$ itself has infinitely many quadratic points -- i.e., points defined over the union of all quadratic extensions of $\mathbb{Q}$. This certainly lives inside $\mathbb{Q}^{\operatorname{ab}}$, so gives infinitely many abelian points.

Now I call a curve $X$ hyperelliptic if it admits a $2:1$ map down to $\mathbb{P}^1$. (I say "I call" because I am not making any genus restrictions and requiring the map to be defined over $\mathbb{Q}$. Standard terminology is taking a little while to catch up to me here...) Now any curve of genus $0$ or $2$ is hyperelliptic, as is any curve of genus one with a rational point. So they all have infinitely many abelian points.

If $E$ is an elliptic curve over $\mathbb{Q}$, then what I'm saying is that if it is given as $y^2 = x^3 + Ax + b$, then take $x$ to be any rational number and extract the square root: that will give you an abelian point. One can see that only finitely many of these quadratic points are torsion points, so we are certainly getting positive rank this way. Do these quadratic points already give infinite rank? I'm not sure (but I feel like I am forgetting something here.) here). [Added: I think I was forgetting what is in Dror Speiser's nice comment below!] Note that here I am -- anemically -- addressing your question a).

A genus one curve without rational points need not be hyperelliptic, and in my paper I construct lots of genus one curves over $\mathbb{Q}$ without elliptic points. This is the key part, actually, because using this I construct curves of every genus $g \geq 4$ over $\mathbb{Q}$ without elliptic abelian points.

This leaves genus $3$, which I was frustratingly unable to deal with in the paper and still can't. (In an appendix, I show that there are genus $3$ curves over some field without points over the maximal abelian extension of that field, unlike in the hyperelliptic cases. So my guess is that this should be possible over $\mathbb{Q}$ as well.)

I didn't think at all about the problem of infinitely versus finitely many abelian points, probably because it cannot be attacked using the methods of my paper. But of course it is interesting too.

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As dke mentioned, I have a paper in which I construct various kinds of varieties $X_{/\mathbb{Q}}$ without abelian points (i.e., with $X(\mathbb{Q}^{\operatorname{ab}}) = \varnothing$). Here is a brief summary:

If $X$ admits a $2:1$ map to a variety $Y$ with infinitely many $\mathbb{Q}$-rational points, then $X$ itself has infinitely many quadratic points -- i.e., points defined over the union of all quadratic extensions of $\mathbb{Q}$. This certainly lives inside $\mathbb{Q}^{\operatorname{ab}}$, so gives infinitely many abelian points.

Now I call a curve $X$ hyperelliptic if it admits a $2:1$ map down to $\mathbb{P}^1$. (I say "I call" because I am not making any genus restrictions and requiring the map to be defined over $\mathbb{Q}$. Standard terminology is taking a little while to catch up to me here...) Now any curve of genus $0$ or $2$ is hyperelliptic, as is any curve of genus one with a rational point. So they all have infinitely many abelian points.

If $E$ is an elliptic curve over $\mathbb{Q}$, then what I'm saying is that if it is given as $y^2 = x^3 + Ax + b$, then take $x$ to be any rational number and extract the square root: that will give you an abelian point. One can see that only finitely many of these quadratic points are torsion points, so we are certainly getting positive rank this way. Do these quadratic points already give infinite rank? I'm not sure (but I feel like I am forgetting something here.) Note that here I am -- anemically -- addressing your question a).

A genus one curve without rational points need not be hyperelliptic, and in my paper I construct lots of genus one curves over $\mathbb{Q}$ without elliptic points. This is the key part, actually, because using this I construct curves of every genus $g \geq 4$ over $\mathbb{Q}$ without elliptic points.

This leaves genus $3$, which I was frustratingly unable to deal with in the paper and still can't. (In an appendix, I show that there are genus $3$ curves over some field without points over the maximal abelian extension of that field, unlike in the hyperelliptic cases. So my guess is that this should be possible over $\mathbb{Q}$ as well.)

I didn't think at all about the problem of infinitely versus finitely many abelian points, probably because it cannot be attacked using the methods of my paper. But of course it is interesting too.