Manin-Drinfeld and constructing a finite morphism with two given ramification points

Fix a smooth projective connected curve $X$ over $\overline{\mathbf{Q}}$ of genus $g\geq 1$ and distinct points $x,y \in X$ such that $x-y$ has infinite order in the Jacobian.

Can we always find a finite morphism $X\to \mathbf{P}^1$ which ramifies at $x$ and $y$?

If not, can we always find $x,y\in X$ such that $x-y$ has infinite order in $\mathrm{Jac}(X)$ and a finite morphism $X\to \mathbf{P}^1$ which ramifies at $x$ and $y$?

Application: If one of the above questions has a positive answer, Belyi's theorem gives the existence of a finite morphism $X\to \mathbf{P}^1$ which ramifies over exactly three points and such that $x$ and $y$ ramify. From this it is easy to see that there exists a subgroup $\Gamma\subset \Gamma(2)$ of finite index such that $\Gamma$ gives a Belyi uniformization of $X$ and such that the Manin-Drinfeld theorem doesn't hold for $\Gamma$.

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The answer is yes (assuming you are not demanding that the map be unramified away from $x$ and $y$). Choose any Belyi map $f: X \to \mathbb{P}^1$. The points $f(x)$ and $f(y)$ are defined over some number field. As part of the proof of his theorem, Belyi shows how to construct finite maps $g: \mathbb{P}^1 \to \mathbb{P}^1$ that are unramified away from $0,1,\infty$ on the target, whose ramification locus on the source contains any finite collection of points defined over a number field, so we can choose $g$ to be ramified at $f(x)$, $f(y)$, 0, 1, $\infty$. The composition $g \circ f$ is what you want.