By what rule or property this expression is equal to six? $$2^{-1} \equiv 6\mod{11}$$
Sorry for very strange question. I want to understand on which algorithm there is a computation of this expression. Similarly interested in why this expression is equal to two?
$$6^{-1} \equiv 2\mod11$$

The notation $2^{-1} \pmod{11}$ means the multiplicative inverse of $2$ modulo $11$, that is, a number $x$ satisfying the equation $2x \equiv 1 \pmod{11}$.  
Since $2 \cdot 6 \equiv 12 \equiv 1 \pmod{11}$, $2^{-1} \equiv 6 \pmod{11}$.  The same calculation shows that $6^{-1} \equiv 2 \pmod{11}$.  
The multiplicative inverse of $6$ exists modulo $11$ since $\gcd(6, 11) = 1$.
We can find the multiplicative inverse of $6$ modulo $11$ using the extended Euclidean algorithm.
\begin{align*}
11 & = 1 \cdot 6 + 5\\
6 & = 1 \cdot 5 + 1\\
5 & = 5 \cdot 1
\end{align*}
We solve for $1$ as a linear combination of $6$ and $11$.
\begin{align*}
1 & = 6 - 5\\
  & = 6 - (11 - 6)\\
  & = 2 \cdot 6 - 11
\end{align*}
Hence, $2 \cdot 6 \equiv 1 \pmod{11} \iff 6^{-1} \equiv 2 \pmod{11}$.
Let's see what happens when we use the extended Euclidean algorithm to solve for $2^{-1}$ modulo $11$.
The multiplicative inverse of $2$ exists modulo $11$ since $\gcd(2, 11) = 1$.
\begin{align*}
11 & = 5 \cdot 2 + 1\\
2 & = 2 \cdot 1
\end{align*}
Solving for $1$ as a linear combination of $2$ and $11$ yields
$$1 = 11 - 5 \cdot 2$$
Thus, $$-5 \cdot 2 \equiv 1 \pmod{11} \iff 2^{-1} \equiv -5 \pmod{11}$$
What does this mean?  It means that for each $t \in \mathbb{Z}$, 
$$2(-5 + 11t) \equiv 1 \pmod{11}$$
In particular, if $t = 1$, we obtain 
$$2 \cdot 6 \equiv 1 \pmod{11} \iff 2^{-1} \equiv  6 \pmod{11}$$
Why is $2(-5 + 11t) \equiv 1 \pmod{11}$ for $t \in \mathbb{Z}$?  Observe that 
$$2(-5 + 11t) \equiv -10 + 22t \equiv 1 - 11 + 22t \equiv 1 + 11(2t - 1) \equiv 1 \pmod{11}$$
