a) (Solution of M.Mankevich, E.Dovgialo.) Let $\sigma_1 = a + b + c$, $\sigma_2 = ab + bc + ca$, $\sigma_3 = abc$. From the given equalities we have the system
$$\sigma_1 = \sigma_1^2 - 2\sigma_2 \quad (1), \quad \sigma_1^2 = \sigma_1^3 - 3\sigma_1\sigma_2 + 3\sigma_3. \quad (2)$$
From (1) we have $\sigma_1^2 = \sigma_1^3 - 2\sigma_1\sigma_2$. Then from (2) it follows that
$$-2\sigma_1\sigma_2 = -\sigma_1\sigma_2 + 3\sigma_3. \quad (3)$$
Finally, the expression we search is
$$\frac{\sigma_2}{\sigma_3}(\sigma_1 - 2) = \frac{\sigma_1\sigma_2 - 2\sigma_2}{\sigma_3} = [\text{in view of (3)}] = \frac{3\sigma_3}{\sigma_3} = 3.$$

b) It is easy to verify, that $a = \frac{1}{2}$, $b = \frac{2+\sqrt{6}}{4}$, $c = \frac{2-\sqrt{6}}{4}$ satisfy the given system.