For certain real values of $a, b, c,$ and $d,$ the equation $x^4+ax^3+bx^2+cx+d=0$ has four non-real roots.  The product of two of these roots is $13+i$ and the sum of the other two roots is $3+4i,$ where $i=\sqrt{-1}.$  Find $b.$
1. **Identify the roots and their properties:**
   Given that the polynomial \(x^4 + ax^3 + bx^2 + cx + d = 0\) has four non-real roots, and the roots must come in conjugate pairs. Let the roots be \( \alpha, \overline{\alpha}, \beta, \overline{\beta} \).

2. **Given conditions:**
   - The product of two of the roots is \(13 + i\).
   - The sum of the other two roots is \(3 + 4i\).

3. **Conjugate pairs:**
   Since the roots come in conjugate pairs, if \( \alpha \) and \( \overline{\alpha} \) are one pair, and \( \beta \) and \( \overline{\beta} \) are the other pair, then:
   \[
   \alpha + \overline{\alpha} = 3 + 4i \quad \text{and} \quad \beta + \overline{\beta} = 3 - 4i
   \]
   \[
   \alpha \overline{\alpha} = 13 + i \quad \text{and} \quad \beta \overline{\beta} = 13 - i
   \]

4. **Sum and product of roots:**
   The sum of all roots is zero (since the coefficient of \(x^3\) is \(a\), and \(a = 0\) for the sum of roots to be zero):
   \[
   (\alpha + \overline{\alpha}) + (\beta + \overline{\beta}) = 0
   \]
   \[
   (3 + 4i) + (3 - 4i) = 6 \neq 0
   \]
   This indicates a mistake in the initial assumption. Instead, we should consider the correct sum and product properties.

5. **Correcting the sum and product:**
   The correct sum of the roots should be:
   \[
   \alpha + \overline{\alpha} + \beta + \overline{\beta} = 0
   \]
   Given the sum of two roots is \(3 + 4i\), the sum of the other two roots must be \(3 - 4i\).

6. **Calculating \(b\):**
   The coefficient \(b\) is the sum of the products of the roots taken two at a time:
   \[
   b = (\alpha \overline{\alpha} + \alpha \beta + \alpha \overline{\beta} + \overline{\alpha} \beta + \overline{\alpha} \overline{\beta} + \beta \overline{\beta})
   \]
   Given:
   \[
   \alpha \overline{\alpha} = 13 + i \quad \text{and} \quad \beta \overline{\beta} = 13 - i
   \]
   We need to find the sum of the products of the roots taken two at a time.

7. **Using Vieta's formulas:**
   The sum of the products of the roots taken two at a time is:
   \[
   b = (\alpha \overline{\alpha} + \beta \overline{\beta}) + (\alpha \beta + \alpha \overline{\beta} + \overline{\alpha} \beta + \overline{\alpha} \overline{\beta})
   \]
   Since \(\alpha \overline{\alpha} = 13 + i\) and \(\beta \overline{\beta} = 13 - i\):
   \[
   b = (13 + i) + (13 - i) = 26
   \]

The final answer is \(\boxed{26}\)