Worked solution: Wiles's modularity proof via the Taylor–Wiles method (1994)

As Wiles explains in the introduction and Chapter 5 of his paper (Wiles 1995, p. 444, p. 448, p. 452–453, and Chapter 5), a semistable elliptic curve $E/\mathbb{Q}$ automatically satisfies the local hypotheses of his lifting theorem at $p = 3$, except possibly when $\overline{\rho}_{E,3}$ is reducible (or reducible over $\mathbb{Q}(\sqrt{-3})$). If both $\overline{\rho}_{E,3}$ and $\overline{\rho}_{E,5}$ were reducible, $E$ would yield a rational point on the modular curve $X_0(15)$ (or a closely related curve), and the known classification of rational points on $X_0(15)$ shows that every such semistable curve is already modular.

When $\overline{\rho}_{E,3}$ is reducible and $\overline{\rho}_{E,5}$ is irreducible, Wiles considers the twisted modular curve $X(\overline{\rho}_{E,5}, \overline{\rho}_{E,3})$ parameterizing elliptic curves $E'$ whose $5$-torsion agrees with $E[5]$ and whose $3$-torsion has a prescribed irreducible image. Because this moduli curve has genus $0$ with rational points, Hilbert's irreducibility theorem produces a semistable elliptic curve $E'/\mathbb{Q}$ such that $\overline{\rho}_{E',5} \cong \overline{\rho}_{E,5}$ while $\overline{\rho}_{E',3}$ is irreducible (Wiles 1995, p. 448 and Chapter 5).

Applying the $p = 3$ modularity lifting theorem (Theorem 0.3) to $E'$ proves that $E'$ is modular. Since $E'$ is modular, its mod-$5$ representation $\overline{\rho}_{E',5}$ comes from a modular form—and because $\overline{\rho}_{E,5} \cong \overline{\rho}_{E',5}$, the mod-$5$ representation of $E$ is now known to be modular! Wiles can then apply his modularity lifting theorem a second time, now with $p = 5$, to conclude that $E$ is modular (Wiles 1995, Theorem 0.4, p. 448).