Find unknown digits $a$ and $b$ such that number $\overline{a783b}$ is divisible with $56$
To find the unknown digits \(a\) and \(b\) such that the number \(\overline{a783b}\) is divisible by \(56\), we need to ensure that it is divisible by both \(8\) and \(7\) (since \(56 = 8 \times 7\)).

1. **Divisibility by 8**:
   For a number to be divisible by \(8\), its last three digits must be divisible by \(8\). Therefore, we need the number \(83b\) to be divisible by \(8\).

   Let's test the possible values of \(b\) (from \(0\) to \(9\)):
   \[
   \begin{aligned}
   &830 \div 8 = 103.75 \quad \text{(not divisible)} \\
   &831 \div 8 = 103.875 \quad \text{(not divisible)} \\
   &832 \div 8 = 104 \quad \text{(divisible)} \\
   &833 \div 8 = 104.125 \quad \text{(not divisible)} \\
   &834 \div 8 = 104.25 \quad \text{(not divisible)} \\
   &835 \div 8 = 104.375 \quad \text{(not divisible)} \\
   &836 \div 8 = 104.5 \quad \text{(not divisible)} \\
   &837 \div 8 = 104.625 \quad \text{(not divisible)} \\
   &838 \div 8 = 104.75 \quad \text{(not divisible)} \\
   &839 \div 8 = 104.875 \quad \text{(not divisible)}
   \end{aligned}
   \]
   The only value of \(b\) that makes \(83b\) divisible by \(8\) is \(b = 2\).

2. **Divisibility by 7**:
   Now, we need the number \(\overline{a7832}\) to be divisible by \(7\). We can use modular arithmetic to check this:
   \[
   a7832 \equiv 0 \pmod{7}
   \]
   We can break this down:
   \[
   a \cdot 10000 + 7000 + 800 + 30 + 2 \equiv 0 \pmod{7}
   \]
   Simplifying each term modulo \(7\):
   \[
   10000 \equiv 4 \pmod{7} \quad (\text{since } 10000 = 7 \cdot 1428 + 4)
   \]
   \[
   7000 \equiv 0 \pmod{7}
   \]
   \[
   800 \equiv 1 \pmod{7} \quad (\text{since } 800 = 7 \cdot 114 + 2)
   \]
   \[
   30 \equiv 2 \pmod{7} \quad (\text{since } 30 = 7 \cdot 4 + 2)
   \]
   \[
   2 \equiv 2 \pmod{7}
   \]
   Combining these:
   \[
   4a + 0 + 1 + 2 + 2 \equiv 0 \pmod{7}
   \]
   \[
   4a + 5 \equiv 0 \pmod{7}
   \]
   Solving for \(a\):
   \[
   4a \equiv -5 \pmod{7}
   \]
   \[
   4a \equiv 2 \pmod{7} \quad (\text{since } -5 \equiv 2 \pmod{7})
   \]
   To solve \(4a \equiv 2 \pmod{7}\), we multiply both sides by the modular inverse of \(4\) modulo \(7\). The inverse of \(4\) modulo \(7\) is \(2\) (since \(4 \cdot 2 \equiv 1 \pmod{7}\)):
   \[
   a \equiv 2 \cdot 2 \pmod{7}
   \]
   \[
   a \equiv 4 \pmod{7}
   \]
   Therefore, \(a = 4\).

Conclusion:
The digits \(a\) and \(b\) that make the number \(\overline{a783b}\) divisible by \(56\) are \(a = 4\) and \(b = 2\).

The final answer is \(\boxed{a = 4, b = 2}\)