Solution:
Since $y_{1} + y_{2} = 1$ and $y_{1} y_{2} = 6a$, we have
$$\begin{gathered} \left(x^{2} - y_{1} x + a\right)\left(x^{2} - y_{2} x + a\right) \\ = x^{4} - \left(y_{1} + y_{2}\right) x^{3} + \left(2a + y_{1} y_{2}\right) x^{2} - a\left(y_{1} + y_{2}\right) x + a^{2} = f(x) \end{gathered}$$

b) The roots of $f(x) = 0$ are the roots $x_{1}, x_{2}$ and $x_{3}, x_{4}$ of the equations
$$x^{2} - y_{1} x + a = 0 \text{ and } x^{2} - y_{2} x + a = 0$$
respectively. It is easy to see that $x_{1}, x_{2}, x_{3}$ and $x_{4}$ are real, distinct and positive exactly when the following three conditions are simultaneously satisfied:

1) $a \neq 0, y_{1} \neq y_{2}$ are real $\Longleftrightarrow a \neq 0, D = 1 - 24a > 0 \Longleftrightarrow a \neq 0$, $a < \frac{1}{24}$;

2) $D_{1} = y_{1}^{2} - 4a > 0$ and $D_{2} = y_{2}^{2} - 4a > 0 \Longleftrightarrow y_{1} > 10a$ and $y_{2} > 10a$;

3) $x_{1} + x_{2} = y_{1} > 0, x_{1} x_{2} = a > 0, x_{3} + x_{4} = y_{2} > 0$ and $x_{3} x_{4} = a > 0$ $\Longleftrightarrow y_{1} > 0, y_{2} > 0$ and $a > 0$.

These conditions are equivalent to $0 < a < \frac{1}{24}, y_{1} > 10a$ and $y_{2} > 10a$, which is the same as $0 < a < \frac{1}{24}, g(10a) > 0$ and $10a < \frac{1}{2} \Longleftrightarrow 0 < a < \frac{1}{24}$, $4a(25a - 1) > 0$ and $a < \frac{1}{20}$.

Hence the required values of $a$ are $a \in \left(\frac{1}{25}, \frac{1}{24}\right)$.