Calculate the multiplicative inverse of $7$ in $\mathbb{Z}_{12}$ I'm not sure how to do this but I gave it a try:
I know that $7$ must have an inverse because $\text{ gcd }(12,7)=1$
Now, on this, use euclidean algorithm. We have $$7x \equiv 1 \text{ mod } 12$$
$$x \equiv 7^{-1}(\text{ mod } 12)$$
So
$$12 = 7 \cdot 1+5$$
$$7=5 \cdot 1+2$$
$$5=2\cdot 2+1$$
Now we can stop this here and write:
$$1=5-4$$
And now I cannot replace $4$ because there isn't any $4$ in the equations above : /
What to do? There is another, easier way?

You have
\begin{align}
\color{red}{12}&=\color{red}{7}\cdot1+\color{red}{5}\\
\color{red}{7} &=\color{red}{5}\cdot1+\color{red}{2}\\
\color{red}{5} &=\color{red}{2}\cdot2+\color{green}{1}
\end{align}
Then
$$
1=\color{red}{5}-\color{red}{2}\cdot2=
\color{red}{5}-2(\color{red}{7}-\color{red}{5})=
-2\cdot\color{red}{7}+3\cdot\color{red}{5}=
-2\cdot\color{red}{7}+3(\color{red}{12}-\color{red}{7})=
-5\cdot\color{red}{7}+3\cdot\color{red}{12}
$$
