$13 f(x)=x^{4}-x^{3}+8 a x^{2}-a x+a^{2}$. Find all real values of the parameter $a$ such that the equation $f(x)=0$ has four distinct positive roots.
Let $y_{1}+y_{2}=1$, $y_{1} y_{2}=6 a$, i.e., they are the roots of $g(x)=x^{2}-x+6 a=0$, then
$$
f(x)=\left(x^{2}-y_{1} x+a\right)\left(x^{2}-y_{2}+a\right).
$$

Let $x_{1}, x_{2}$ and $x_{3}, x_{4}$ be the roots of $x^{2}-y_{1} x+a=0$ and $x^{2}-y_{2}+a=0$ respectively.
1) $a>0, y_{1} \neq y_{2} \Leftrightarrow 00, D_{2}=y_{2}^{2}-4 a>0 \Leftrightarrow y_{1}>10 a, y_{2}>$
$10 a$
$$
\begin{aligned}
\text { 3) } & x_{1}+x_{2}=y_{1}>0, x_{1} x_{2}=a>0, x_{3}+x_{4}=y_{2}>0, \\
x_{3} x_{4}= & a>0 \Leftrightarrow y_{1}, y_{2}, a>0. \\
\Leftrightarrow & 00,10 a<\frac{1}{2} \Leftrightarrow a \in\left(\frac{1}{25}, \frac{1}{24}\right).
\end{aligned}
$$