Finding the inverse of a number under a certain modulus How does one get the inverse of 7 modulo 11?
I know the answer is supposed to be 8, but have no idea how to reach or calculate that figure.
Likewise, I have the same problem finding the inverse of 3 modulo 13, which is 9.

Here's an illustration of finding the multiplicative inverse of $37 \bmod 100$ using the extended Euclidean algorithm. (I used bigger numbers for this example so that the relationships are a little clearer).
On each line, $n=100s+37t$. We start the table with two lines giving $n=100$ and $n=37$ in the obvious way. Then $q$ gives the rounded-down ratio of the current and previous value of $n$.  Using this, the next line is calculated by subtracting $q$ copies of the current line entries from the  previous line entries.
$$\begin{array}{c|c|c|c}
n & s & t & q \\ \hline
100 & 1 & 0 &  \\
37 & 0 & 1 & 2 \\
26 & 1 & -2 & 1 \\
11 & -1 & 3 & 2 \\
4 & 3 & -8 & 2 \\
3 & -7 & 19 & 1 \\
1 & 10 & \color{red}{-27} & 3 \\
\end{array}$$
On the last line, $1 = 10\times 100 + (-27)\times 37$, so $\color{red}{-27}\times 37 \equiv 1 \bmod 100$ 
Bringing the result positive, $-27 \equiv \color{red}{73} \bmod 100$. And $37 \times 73 = 2701 \equiv 1 \bmod 100 \quad \checkmark$.
As you can perhaps see, you don't actually need to calculate the $s$ values at all to get the modular inverse. I left them in to help understand the table.
The corresponding tables for your particular questions:
$$\begin{array}{c|c|c|c}
n & s & t & q \\ \hline
11 & 1 & 0 &  \\
7 & 0 & 1 & 1 \\
4 & 1 & -1 & 1 \\
3 & -1 & 2 & 1 \\
1 & 2 & \color{red}{-3} & 3 \\
\end{array}$$
and $7^{-1} \equiv -3 \equiv \color{red}8  \bmod 11$. $\quad 7\times 8 = 56 = 55+1\quad \checkmark$
$$\begin{array}{c|c|c|c}
n & s & t & q \\ \hline
13 & 1 & 0 &  \\
3 & 0 & 1 & 4 \\
1 & 1 & \color{red}{-4} & 3 \\
\end{array}$$
and $3^{-1} \equiv -4 \equiv \color{red}9  \bmod 13$.  $\quad 3\times 9 = 27 = 26+1\quad \checkmark$
