How many four-digit positive integers are there that contain the digit $3$ and are divisible by $5$? 
How many four-digit positive integers are there that contain the digit $3$ and are divisible by $5$?

The answer is: 
the  number of four-digit integers that are divisible by $5\;-\;$ the number of four-digit integers that are divisible by $5$ and not contain the digit $3$
So, $9\cdot10\cdot10\cdot2-8\cdot 9\cdot 9\cdot 2=1800-1296=504\tag{*}$
I know that, but when I tried to solve this problem with an other way I got a different result.
Four digit integers : $\overline{xyzw}$
Suppose $\overline{xyzw}$ contain at least one digit equal $3$
So, $x = \{1, \ldots ,9\}, y = 3, z = \{0, \ldots, 9\}, w = \{0,5\}$
or $x = \{1, \ldots, 9\}, y = \{0, \ldots, 9\}, z = 3, w = \{0,5\}$
or $x = 3, y=\{0, \ldots, 9\}, z = \{0, \ldots, 9\}, w = \{0,5\}$
The number of all $\overline{xyzw}$ that must be divisible by $5$ and contain $3$ is: $9\cdot 10\cdot 2+9\cdot 10\cdot 2+10\cdot 10\cdot 2=180+180+200=560\tag{**}$
but (*) contradicts (**), so where is the mistake?

You double counted, for example $3335$ appears in all your sum.
You can use inclusion-exclusion to adjust your count.
\begin{align}
&|A|+|B|+|C|-|A\cap B|-|A \cap C|-|B \cap C| + |A \cap B \cap C|\\
&=560 - 18 - 20-20+2\\
&=504
\end{align}
