Recall Euler's theorem: if [tex]\gcd(a,n) = 1[/tex], then
[tex]a^{\phi(n)} \equiv 1 \pmod n[/tex]
where [tex]\phi[/tex] is Euler's totient function.
We have [tex]\gcd(9,32) = 1[/tex] - in fact, [tex]\gcd(9,32^k)=1[/tex] for any [tex]k\in\Bbb N[/tex] since [tex]9=3^2[/tex] and [tex]32=2^5[/tex] share no common divisors - as well as [tex]\phi(9) = 6[/tex].
Now,
[tex]37^{32} = (1 + 36)^{32} \\\\ ~~~~~~~~ = 1 + 36c_1 + 36^2c_2 + 36^3c_3+\cdots+36^{32}c_{32} \\\\ ~~~~~~~~ = 1 + 6 \left(6c_1 + 6^3c_2 + 6^5c_3 + \cdots + 6^{63}c_{32}\right) \\\\ \implies 32^{37^{32}} = 32^{1 + 6(\cdots)} = 32\cdot\left(32^{(\cdots)}\right)^6[/tex]
where the [tex]c_i[/tex] are positive integer coefficients from the binomial expansion. By Euler's theorem,
[tex]\left(32^{(\cdots)\right)^6 \equiv 1 \pmod9[/tex]
so that
[tex]32^{37^{32}} \equiv 32\cdot1 \equiv \boxed{5} \pmod9[/tex]
