Shortcut to find remainder (mod) of a number to a negative power? How would one find the remainder to a question of the type $a^{-x} / b$ really fast? 

Came across this situation while solving the remainder of $12^{107} / 37$ 

Using Fermat's little theorem we get 
$12^{108} \mod 37 = 1$
which would mean the question is nothing, but the remainder of $(12^{-1} \times 12^{108} ) / 37 \implies 12^{-1} \times 1 / 37$ (Basic remainder theorem) 

Is there an easy and fast method to solving this question from this step? 

I did manage to solve it but with 4 additional steps, where I basically got  the numerator to a 3-digit value applying  Fermat's and basic remainder theorem  multiple times. But in a competitive time crunch exam situation, I do not think my method is ideal, there must be a way to easily solve $12^{-1} \mod 37$ fast.
It is trivial to invert a number that divides modulus $\pm 1,\,$ e.g. here $\,12\mid 37-1,\,$ i.e $\,3\cdot 12 = 37-1\,$ so $\,{\rm mod}\ 17\!:\,\ 3\cdot 12\equiv -1\,$ so $\,12^{-1}\equiv -3.\ $  Said in fraction language

$$\dfrac{1}{12}\equiv \dfrac{-36}{12} \equiv \dfrac{-3}{1}$$

This is essentially an optimzation of the extended Euclidean algorithm (when it terminates in one or two steps).  There are other ad-hoc methods that are best expressed in terms of fraction arithmetic (e.g. using factorizations, cancellation, etc), e.g. see some of my prior posts using modular fractions
What you want is the answer to 
$$a\equiv 12^{-1}\pmod{37}$$
or equivalently, on multiplying through by $12$ to
$$12a\equiv 12\cdot12^{-1}\equiv1\pmod{37}$$
where $a$ is the inverse of $12$ modulo $37$.

Now by Fermat's Little Theorem we have:
$$12^{37-1}=12^{36}\equiv1\pmod{37}$$
On noting $12a\equiv 12^{36}\pmod{37}$ we have, on dividing through by $12$, that
$$a\equiv 12^{35}=(12^{5})^{7}=248832^7\equiv 7^{7}=823543\equiv34\equiv-3\pmod{37}$$
So $a\equiv-3$ is the answer.