Let $\{a, b, c, d, e, f, g, h\}$ be a permutation of $\{1, 2, 3, 4, 5, 6, 7, 8\}$. What is the probability that $\overline{abc} +\overline{def}$ is even?
1. **Identify the condition for the sum to be even:**
   For the sum $\overline{abc} + \overline{def}$ to be even, the last digits $c$ and $f$ must sum to an even number. This is because the sum of two numbers is even if and only if their units digits sum to an even number.

2. **Determine the possible values for $c$ and $f$:**
   The set $\{1, 2, 3, 4, 5, 6, 7, 8\}$ contains four even numbers $\{2, 4, 6, 8\}$ and four odd numbers $\{1, 3, 5, 7\}$.

3. **Calculate the number of ways to pick two numbers such that their sum is even:**
   - **Case 1: Both numbers are even.**
     There are $\binom{4}{2}$ ways to choose 2 even numbers from 4 even numbers.
     \[
     \binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4 \times 3}{2 \times 1} = 6
     \]
   - **Case 2: Both numbers are odd.**
     There are $\binom{4}{2}$ ways to choose 2 odd numbers from 4 odd numbers.
     \[
     \binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4 \times 3}{2 \times 1} = 6
     \]

4. **Calculate the total number of favorable outcomes:**
   The total number of ways to pick two numbers such that their sum is even is the sum of the two cases:
   \[
   6 \text{ (even)} + 6 \text{ (odd)} = 12
   \]

5. **Calculate the total number of possible outcomes:**
   The total number of ways to pick any two numbers from the set of 8 numbers is:
   \[
   \binom{8}{2} = \frac{8!}{2!(8-2)!} = \frac{8 \times 7}{2 \times 1} = 28
   \]

6. **Calculate the probability:**
   The probability that $\overline{abc} + \overline{def}$ is even is the ratio of the number of favorable outcomes to the total number of possible outcomes:
   \[
   \frac{12}{28} = \frac{3}{7}
   \]

Conclusion:
\[
\boxed{\frac{3}{7}}
\]