Find the inverse of the equivalence class I have to check if the equivalence class has an inverse(without calculations). If yes, I have to find it.
$$[7] \in \mathbb{Z}_{36}$$
We know that $[a] \in \mathbb{Z}_m$ has an inverse $\Leftrightarrow (a,m)=1$.
In this case, knowing that $(7,36)=1$ we conclude that $[7]$ has an inverse. 
$$\text{So there is a } x \text{ such that } [7][x]=[1] \text{ in } \mathbb{Z}_{36}.$$
But how can I find this $x$? Do I have to check all integers that are in $\mathbb{Z}_{36}$, that means all integers in $\{0,1,...,35\}$?
 A: Depends on whether you are supposed to use "general" methods or not. If not, we have $5\cdot 7=35\equiv -1\pmod{36}$, so the equivalence class of $-5$ is the inverse. We could also call it the equivalence class of $31$. 
A: You can use the Extended Euclidean algorithm to compute integers
$u, v$ such that
$$
    a \cdot u+ m \cdot v = \gcd(a, m) = 1
$$
If follows that
$$
a \cdot u = 1 \pmod{m}
$$
which means that $[u]$ is the inverse to $[a]$ in $\mathbb Z_m$.
A: A way to find this inverse is to compute $[7^{\phi(36)-1}] = [7^{11}] = [31]$, using Euler's theorem, which states that $[a^{\phi(m)}] = [1]$ in $\mathbb{Z}_m$.
A: Hint $\ {\rm mod}\ 36\!:\,\ 7x\equiv 1\,\Rightarrow\, x\,\equiv\,  \dfrac{1}7 \,\equiv\, \dfrac{-35}7 \,\equiv\, -5.\,$ 

Or $\bmod 6^2\!:\,\ \dfrac{1}{1+6}= 1-6\ $ by simpler multiples or Hensel lifting or Newton's method.

More generally we can employ the Extended Euclidean Algorithm and related techniques to compute modular inverses. A few worked examples using a handful of methods are  here and here and here.
