2.4. By using the condition $a+b+c=1$ and the inequality between arithmetic and geometric means, we have that
$$\frac{a}{a+b^2} = \frac{a}{a(a+b+c)+b^2} = \frac{a}{a^2+b^2+ab+ac} \le \frac{a}{2ab+ab+ac} = \frac{1}{3b+c}.$$
By applying the inequality between harmonic and arithmetic means, it follows that
$$\frac{4}{\frac{3}{b} + \frac{1}{c}} = \frac{4}{\frac{1}{b} + \frac{1}{b} + \frac{1}{b} + \frac{1}{c}} \le \frac{b+b+b+c}{4} = \frac{3b+c}{4},$$
i.e.
$$\frac{a}{a+b^2} \le \frac{1}{3b+c} \le \frac{1}{16} \left( \frac{3}{b} + \frac{1}{c} \right).$$
Analogously,
$$\frac{b}{b+c^2} \le \frac{1}{16} \left( \frac{3}{c} + \frac{1}{a} \right) \quad \text{and} \quad \frac{c}{c+a^2} \le \frac{1}{16} \left( \frac{3}{a} + \frac{1}{b} \right).$$
Finally, by adding the inequalities above, we have that
$$\frac{a}{a+b^2} + \frac{b}{b+c^2} + \frac{c}{c+a^2} \le \frac{1}{16} \left( \frac{4}{a} + \frac{4}{b} + \frac{4}{c} \right) = \frac{1}{4} \left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \right).$$

Let the roots of $x^4 + a x^2 + b = 0$ be $x_1$, $x_2$, $x_3$, $x_4$.

Let us factor the quartic as follows:
Let $y = x^2$, so the equation becomes $y^2 + a y + b = 0$.
Let the roots of this quadratic be $y_1$ and $y_2$.
Then the roots of the quartic are $x_1 = \sqrt{y_1}$, $x_2 = -\sqrt{y_1}$, $x_3 = \sqrt{y_2}$, $x_4 = -\sqrt{y_2}$.

The sum of squares of all solutions is:
$x_1^2 + x_2^2 + x_3^2 + x_4^2 = y_1 + y_1 + y_2 + y_2 = 2(y_1 + y_2)$.
We are given $x_1^2 + x_2^2 + x_3^2 + x_4^2 = 32$, so $2(y_1 + y_2) = 32 \implies y_1 + y_2 = 16$.

The product of all solutions is $x_1 x_2 x_3 x_4 = (\sqrt{y_1})(-\sqrt{y_1})(\sqrt{y_2})(-\sqrt{y_2}) = (y_1)(y_2)$.
We are given $x_1 x_2 x_3 x_4 = 4$, so $y_1 y_2 = 4$.

Recall that for the quadratic $y^2 + a y + b = 0$, the sum of roots is $-(a)$ and the product is $b$.
So:
$y_1 + y_2 = -a = 16 \implies a = -16$
$y_1 y_2 = b = 4$

Therefore, $a = -16$ and $b = 4$.