Solution:
The first condition of the problem is equivalent to the assertion that the equation $x^{3} + a x + b = 0$ has a double real root $x_{1} \neq 0$ and a simple real root $x_{2} \neq 0$, where $x_{2} \neq x_{1}$. Therefore
$$x^{3} + a x + b = (x - x_{1})^{2}(x - x_{2})$$
We also have $\angle ACB = 90^{\circ}$, where $A = (x_{1}, 0)$, $B = (x_{2}, 0)$, $C = (0, b)$ and $x_{1} x_{2} < 0$. Hence $AO \cdot BO = CO^{2}$, i.e. $-x_{1} x_{2} = b^{2}$. Since $b = -x_{1}^{2} x_{2}$ and $x_{1}, x_{2} \neq 0$, we get $x_{1}^{3} x_{2} = -1$. On the other hand, we have $2 x_{1} + x_{2} = 0$ and therefore $2 x_{1} = -x_{2} = \frac{1}{x_{1}^{3}}$. Hence $x_{1} = \pm \frac{1}{\sqrt[4]{2}}$ and $x_{2} = \mp \sqrt[4]{8}$. Then
$a = x_{1}^{2} + 2 x_{1} x_{2} = -\frac{3}{\sqrt{2}}$ and $b = -x_{1}^{2} x_{2} = \pm \sqrt[4]{2}$. The above arguments imply as well that these two values of $b$ are solutions indeed.