The Galois action on torsion points lands in GSp(2g, Z/n) whose size grows rather rapidly with g and n. So the kernel of this rep will typically have quite enormous degree.

Yes, once you know that $\rho$ is automorphic, it follows that $L(s, \rho \otimes \chi)$ has good analytic properties for all Dirichlet characters $\chi$, and more generally so does $L(s, \rho \otimes \sigma)$ for any automorphic representation $\sigma$ of $GL_m$ (any $m$).

It might interest the OP that the computer algebra systems "Sage" and "Magma" have practical, workable implementations of rigorous algebraic-number computation (including square roots!) using exactly this method. (Minor detail: I think the implementations use rectangular boxes, not circles, to isolate the roots, since this makes the computations easier).

Appendix B of Rubin's book "Euler Systems" is a good reference for this.

Which analogy are you referring to?

Bellaiche's Clay summer school lectures on the Bloch--Kato conj are also a good starting point (perhaps a good warmup before you tackle Fontaine-Perrin-Riou).

I don't think the assertion about $k\alpha = \beta mod 1$ is equivalent to $\mathbb{Q}(\alpha) = \mathbb{Q}(\beta)$. Clearly $\alpha = \sqrt{2} + \sqrt{3}$ and $\beta = 2\sqrt{2} + 3\sqrt{3}$ generate the same extension, but I don't think that $k\alpha = \beta \bmod 1$ is solvable.

Regarding the use of the j-invariant: perhaps it might be possible to avoid circularity by using a weaker form of the valence formula to establish the necessary properties of the j-invariant, and then looping back round to establish the full valence formula from this. E.g. perhaps one could show by other, more elementary means that $j$ has no zeroes on the boundary of $\mathcal{D}$ except the two at the corners, and then one can integrate around a much simpler contour.