Which integers have order 6 (mod 37)? Which integers have order $6\ (\text {mod}\ 37)$?
I have tried expanding $x^6-1=0\ (\text {mod}\ 37)$ equation, i tried expanding it as $x^3-1$ and $x^3+1$ and I expanded further and tried to expand as adding $37$ to $x^6-1$ does alter as $\text {mod}\ 37$ is zero , but I could lead further.
HINT:

Try to use Fermats little theorem.

EDIT: By the above named theorem $x^{36}=(x^6)^6\equiv 1 \ \ \bmod 37$. Not surprisingly $x\equiv_{37}1$ is your first solution. $2^6=64\equiv_{37} 27$ so $27$ goes on the list. $3^6=3^2\cdot 3^2\cdot 3=81\cdot 3^2\equiv_{37}7\cdot 3^2\equiv_{37}26$ so $26$ goes also on the list. $4^6=(2^6)^2\equiv_{37}=27^2\equiv_{37}26$. And so on... You will be able to use already known information as soon as $x$ is a composite.
Look for a generator, which is any number $g$ such that $g^{12}\not\equiv 1$ and $g^{18}\not\equiv 1$. Here you get lucky, because the first number you try, $2$, is a generator. So now it's just $2^6$ and $2^{30}$.