12.2. 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

$$x^{3}+a x+b=\left(x-x_{1}\right)^{2}\left(x-x_{2}\right)$$

We also have $\Varangle A C B=90^{\circ}$, where $A=\left(x_{1}, 0\right), B=\left(x_{2}, 0\right), C=(0, b)$ and $x_{1} x_{2}<0$. Hence $A O . B O=C O^{2}$, 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.