12. (20 points) Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{0}=\frac{1}{k}, a_{n}=a_{n-1}+\frac{1}{n^{2}} a_{n-1}^{2}\left(k \in \mathbf{Z}_{+}\right) \text {. }
$$

If for all $n \in \mathbf{Z}_{+}$, we have $a_{n}<1$, find the range of values for $k$.
12. When $k=1$, i.e., $a_{0}=1$, by mathematical induction we know
$$
a_{n}=n+1\left(n \in \mathbf{Z}_{+}\right) \text {. }
$$

When $k=2$, we have
$$
\begin{array}{l}
a_{n}-a_{n-1}=\frac{1}{n^{2}} a_{n-1}^{2}\left(n \in \mathbf{Z}_{+}\right) \\
\Rightarrow a_{n}-a_{0}=\sum_{k=1}^{n} \frac{a_{k-1}^{2}}{k^{2}} .
\end{array}
$$

Clearly, $\left\{a_{n}\right\}$ is a monotonically increasing sequence.
$$
\begin{array}{l}
\text { Hence } a_{n}-a_{0}>a_{1}^{2} \sum_{k=2}^{n} \frac{1}{k^{2}}+a_{0}^{2} \\
>\left(\frac{1}{2}-\frac{1}{n+1}\right) \frac{9}{16}+\frac{1}{4} \\
\Rightarrow a_{n}>\frac{1}{32}+\frac{3}{4}-\frac{9}{16(n+1)} \\
=\frac{33}{32}-\frac{9}{16(n+1)} .
\end{array}
$$

When $\frac{9}{16(n+1)}1$.
When $k \geqslant 3$, we have
$$
\begin{array}{l}
a_{n}-a_{n-1}=\frac{1}{n^{2}} a_{n-1}^{2}1 \Rightarrow a_{n}<1 .
\end{array}
$$

Therefore, when $k \geqslant 3$, we have $a_{n}<1\left(n \in \mathbf{Z}_{+}\right)$.