10. (20 points) Given that $\left\{a_{n}\right\}$ is a bounded sequence, and
$$
a_{1}=a, a_{n+1}=3 a_{n}^{3}-7 a_{n}^{2}+5 a_{n}\left(n \in \mathbf{Z}_{+}\right) \text {. }
$$

Find the range of values for $a$.
10. From $\frac{a_{n+1}}{a_{n}}=3 a_{n}^{2}-7 a_{n}+5>0$ always holds, we know that $a_{n}$ and $a$ have the same sign.
(1) If $a5^{n-1}|a| .
\end{array}
$$

For any positive integer $M$, we only need to take
$$
N=\max \left\{1,\left[\log _{5} \frac{M}{|a|}\right]+2\right\} \text {, }
$$

then when $n>N$, $\left|a_{n}\right|>M$, which contradicts that $\left\{a_{n}\right\}$ is a bounded sequence.
(2) If $a>\frac{4}{3}$, then $a_{n}\left(n \in \mathbf{Z}_{+}\right)$ is positive.
By $a_{n+1}-a_{n}=a_{n}\left(a_{n}-1\right)\left(3 a_{n}-4\right)>\frac{1}{3} a_{n}$
$$
\Rightarrow a_{n+1}>\frac{4}{3} a_{n} \Rightarrow a_{n}>\left(\frac{4}{3}\right)^{n-1} a \text {. }
$$

For any positive integer $M$, we only need to take
$$
N=\max \left\{1,\left[\log _{\frac{4}{3}} \frac{M}{|a|}\right]+2\right\} \text {, }
$$

then when $n>N$, $\left|a_{n}\right|>M$, which contradicts that $\left\{a_{n}\right\}$ is a bounded sequence.
(3) If $0 \leqslant a \leqslant \frac{4}{3}$, let
$$
\begin{aligned}
f(x) & =3 x^{3}-7 x^{2}+5 x \\
\Rightarrow & f^{\prime}(x)=9 x^{2}-14 x+5 \\
& =(9 x-5)(x-1) .
\end{aligned}
$$

Thus, $f(x)$ is monotonically increasing on the interval $\left(0, \frac{5}{9}\right)$, monotonically decreasing on the interval $\left(\frac{5}{9}, 1\right)$, and monotonically increasing on the interval $\left(1, \frac{4}{3}\right)$.
Notice that, $f(0)=0, f(1)=1$,
$$
f\left(\frac{5}{9}\right)=\frac{275}{243}<f\left(\frac{4}{3}\right)=\frac{4}{3} \text {. }
$$

Hence, $f(x) \in\left[0, \frac{4}{3}\right]$.
By mathematical induction, it is easy to know that $a_{n} \in\left[0, \frac{4}{3}\right]$.
Therefore, $\left\{a_{n}\right\}$ is a bounded sequence.
In summary, $0 \leqslant a \leqslant \frac{4}{3}$.