How do I solve $32x \equiv 12 \pmod {82}$?

Hint :you can do like this $$\quad{8x \equiv 3 \pmod {41}\\ 8x \equiv 3+41 \pmod {41}\\8x \equiv 44 \pmod {41} \div4 \\ 2x \equiv 11 \pmod {41}\\2x \equiv 11+41 \pmod {41}\\2x \equiv 52 \pmod {41}\div 2\\x \equiv 26 \pmod {41}\\x=41q+26}$$

You just have to find the inverse of $8$ modulo $41$.

The general method uses the extended Euclidean algorithm, but in the particular case, it's much simpler: from $5\cdot 8=40\equiv -1\mod 41$, you get at once that $8^{-1}\equiv -5\mod 41$, so $$x\equiv -5\cdot 3=-15\equiv 26\mod 41.$$