Let $r>2$ and let $b_1,b_2,\ldots,b_r$ be in $\mathbf{P}^1(\mathbf{Q})$. Let $B$ be the divisor $$B:= \sum [b_i].$$ We consider this data to be fixed. For $d>1$, we define $\textrm{Ell}(b_1,b_2,\ldots,b_r,d)$ as the set of (isomorphism classes of) elliptic curves $E$ over $\mathbf{Q}$ that admit a finite morphism $f:E\longrightarrow \mathbf{P}^1_\mathbf{Q}$ of degree $d$ which is etale outside $\{b_1,b_2,\ldots,b_r\} \subset \mathbf{P}^1(\mathbf{Q})$.

Question 1. Let $E$ be in $\textrm{Ell}(b_1,b_2,\ldots,b_r,d)$ and choose a finite morphism $f:E\longrightarrow \mathbf{P}^1_\mathbf{Q}$ of degree $d$ which is etale outside $\{b_1,b_2,\ldots,b_r\} \subset \mathbf{P}^1(\mathbf{Q})$. Let $X$ be the analytification of $E_\mathbf{C}$. There exists a $\tau$ in the complex upper half plane such that $X = \mathbf{C}/\mathbf{Z}+\tau\mathbf{Z}$. Can we choose $\tau$ (or $q=e^{2\pi i \tau}$) using the data $(b_1,b_2,\ldots,b_r,d,f)$?

Question 2. It follows from Faltings's theorem that the set $\textrm{Ell}(b_1,b_2,\ldots,b_r,d)$ is finite. Is there a more elementary proof of this?

EDIT: Let me describe how the elliptic curve is given (in the set-up I have in mind).

Let $U$ be an open subscheme of $\mathbf{P}^1_\mathbf{Z}$ with complement $D$. We suppose that the closed subscheme $D$ is a horizontal divisor on $\mathbf{P}^1_\mathbf{Z}$ such that the base change $D_\mathbf{Q}$ equals $B$ defined above. Let $V\longrightarrow U$ be a finite etale morphism, with $V$ connected. Let $g:Y\longrightarrow \mathbf{P}^1_\mathbf{Q}$ be the normalization of $\mathbf{P}^1_\mathbf{Q}$ in the function field of $V$. We make the following extra assumptions:

1. $Y$ has a $\mathbf{Q}$-rational point.

2. The genus of $Y$ equals 1.

So the morphism $f$ arises like this.

I'm actually more interested in the set-up described above without assumptions 1 and 2. I just figured it would be an easy case to start with because it could/should be handled more directly.