jmc

The prime factorization of $2^3\cdot 4^5\cdot 6^7\cdot 8^9$ is $2^{47}\cdot 3^7$. We see that $2^6\equiv 64\equiv -1(\mathrm{mod}13)$, so $$2^{47}\equiv 2^{6\cdot 7+5}\equiv (2^6{)}^7\cdot 2^5\equiv (-1{)}^7\cdot 32\equiv -32\equiv 7(\mathrm{mod}13),$$and $3^7\equiv 2187\equiv 3(\mathrm{mod}13)$, so $2^{47}\cdot 3^7\equiv 7\cdot 3\equiv 21\equiv \boxed{8}(\mathrm{mod}13)$.