So we know the order of 3 must be 256 yet this is too big number to calculate by hand (or calculator) so we use the fact that there exists another primitive root, g, such that 3 is congruent to g^i mod 256.
Then we have 3^256 congruent to g^(i*256) mod 257 where i is between 1 and 255.
I am stuck on how to proceed from here, my notes say: "As 256 is a power of 2, order of 3 is 256 if and only if i is odd."
I don't get this statement; surely g^(i*256) is congruent to 1 mod 257 for any power that is divisible by the order of g so 256 therefore shouldn't i be able to take any value?
As the comments indicate, $i$ must be odd since $257$ is prime, $257-1=256=2^8$, and the multiplicative order of any element must be a power of $2$. To find the order of $3$ is not so hard even by hand. You find that $3^{16}\equiv -8 \mod 257 $, then $3^{64}\equiv -16\mod 257$ and you are practically done.