Solving: Since $f(x)$ is defined on $(0,+\infty)$, we have
$$
\left\{\begin{array}{l}
2 a^{2}+a+1=2\left(a+\frac{1}{4}\right)^{2}+\frac{7}{8}>0, \\
3 a^{2}-4 a+1>0,
\end{array}\right.
$$

we get
$a>1$ or $a<\frac{1}{3}$, and solving $3 a^{2}-4 a+1$, we get $0<a<5$. Combining (1), we know, $0<a<\frac{1}{3}$ or $1<a<5$.
8 Given that $f(x)$ is a decreasing function defined on $(0,+\infty)$, if $f\left(2 a^{2}+a+\right.$ 1) $<f\left(3 a^{2}-4 a+1\right)$ holds, then the range of values for $a$ is $\qquad$.