= Solution
Write $t(\tau)=\Delta(\tau)/\Delta(2\tau)$. For $\gamma=\begin{pmatrix}a&b\\c&d\end{pmatrix}\in\Gamma_0(2)$, the integral <matrix> $\gamma'=\begin{pmatrix}a&2b\\c/2&d\end{pmatrix}$ has <determinant> one and satisfies $2\gamma\tau=\gamma'(2\tau)$. Both discriminants acquire the same factor $(c\tau+d)^{12}$, so $t$ is invariant. It has neither zeros nor <poles> in the half-plane.
Reduction modulo two gives index three, with two <modular cusp> classes, infinity and zero, of <cusp widths> one and two. These can also be found from the orbits of upper triangular <matrices> on primitive columns modulo two. Compactness follows from the finite-index argument in part (a).
At infinity $t=q^{-1}(1+O(q))$, so it has a <simple pole>. At zero use $S\tau=-1/\tau$ and the <modular discriminant> inversion law:
$$
t(S\tau)=2^{12}\frac{\Delta(\tau)}{\Delta(\tau/2)}=2^{12}q_0(1+O(q_0)),\qquad q_0=e^{\pi i\tau}.
$$
Thus zero is a simple zero, not a <pole>, in its width-two <modular cusp> coordinate. The <single-pole criterion for a spherical coordinate> now gives
$$
\boxed{t=j_2:X(\Gamma_0(2))\overset{\sim}{\longrightarrow}\mathbb P^1_{\mathbb C}.}
$$
This <discriminant-ratio coordinate on X0 2> is different from a ratio of two $j$-invariants.
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