How to do this step quickly in Chinese remainder theorem I have 
$ \begin{cases} x \equiv 2 \pmod 3 \\ x \equiv 4 \pmod 7 \\ x \equiv 5 \pmod8 \end{cases} $
and I don't know how to do this quickly in this step:
$56x_1 \equiv 1 \pmod 3 $ implies $x_1 = 2$
The question is, how to find $x_1, x_2, x_3$ fast? In case $x_1$:
instead of multiplying 56 by $2, 3, 4, 5 \ldots$ and do like this $56*2=112$ so $112 \div 3 \approx 37$ thus $112-37*3=1$
What if $x_n$ where $n \ge 1$ is not $2$ but some big number? I know there is the extended Euclidean algorithm.

For the stated problem you have a shortcut giving you 
$x\equiv 4\equiv -3 \bmod 7$
$x\equiv 5\equiv -3 \bmod 8$  
So $x\equiv  -3 \equiv 53 \bmod 56$  
Then since $53 \equiv 2 \bmod 3$ we have $x\equiv 53 \bmod 168$

Your remark about the extended Euclidean algorithm gives the way forward for more complicated cases. You can find the modular inverse of the appropriate modulus $m_1$ in the next modulus $m_2$ and use that to find the appropriate range of $km_1$ values to synchronize the two modular values.
Suppose we are trying to combine
$x\equiv 4 \bmod 21$
$x\equiv 6 \bmod 11$  
Then $21^{-1}\equiv 10^{-1}\equiv 10 \bmod 11$, so to get $4+21k \equiv 6 \bmod 11$ we can set $k=10\cdot (6-4) = 20 \equiv 9 \bmod 11$ so
 $x\equiv 4+21\cdot9 \equiv 193 \bmod (21\cdot11=231)$
