10. Given the sequence $\left\{a_{n}\right\}$ satisfies: $a_{1}=2 t-2(t \in \mathbf{R}$ and $t \neq \pm 1)$,
$$
a_{n+1}=\frac{2\left(t^{n+1}-1\right) a_{n}}{a_{n}+2 t^{n}-2}\left(n \in \mathbf{N}^{*}\right) \text {. }
$$
(1) Find the general term formula of the sequence $\left\{a_{n}\right\}$;
(2) If $t>0$, compare the size of $a_{n+1}$ and $a_{n}$.
(1) According to the problem, $\frac{a_{n+1}}{t^{n+1}-1}=\frac{\frac{2 a_{n}}{t^{n}-1}}{\frac{a_{n}}{t^{n}-1}+2}$. Let $b_{n}=\frac{a_{n}}{t^{n}-1}$,

then $b_{n+1}=\frac{2 b_{n}}{b_{n}+2} \Rightarrow \frac{1}{b_{n+1}}=\frac{1}{b_{n}}+\frac{1}{2}$. And $b_{1}=\frac{a_{1}}{t-1}=2$, so the sequence $\left\{\frac{1}{b_{n}}\right\}$ is an arithmetic sequence with the first term $\frac{1}{2}$ and common difference $\frac{1}{2}$.
Therefore, $\frac{1}{b_{n}}=\frac{n}{2} \Rightarrow b_{n}=\frac{a_{n}}{t^{n}-1}=\frac{2}{n} \Rightarrow a_{n}=\frac{2\left(t^{n}-1\right)}{n}\left(n \in \mathbf{N}^{*}\right)$.
$$
\begin{array}{l}
\text { (2) Since } a_{n+1}-a_{n}=\frac{2\left(t^{n+1}-1\right)}{n+1}-\frac{2\left(t^{n}-1\right)}{n} \\
=2(t-1)\left[\frac{1}{n+1}\left(t^{n}+t^{n-1}+\cdots+t+1\right)-\frac{1}{n}\left(t^{n-1}+t^{n-2}+\cdots+t+1\right)\right] \\
=\frac{2(t-1)}{n(n+1}\left[n\left(t^{n}+t^{n-1}+\cdots+t+1\right)-(n+1)\left(t^{n-1}+t^{n-2}+\cdots+t+1\right)\right] \\
=\frac{2(t-1)}{n(n+1}\left[n t^{n}-\left(t^{n-1}+t^{n-2}+\cdots+t+1\right)\right] \\
=\frac{2(t-1)}{n(n+1}\left[\left(t^{n}-t^{n-1}\right)+\left(t^{n}-t^{n-2}\right)+\cdots+\left(t^{n}-t\right)+\left(t^{n}-1\right)\right] \\
=\frac{2(t-1)^{2}}{n(n+1}\left[t^{n-1}+t^{n-2}(t+1)+\cdots+t\left(t^{n-2}+t^{n-3}+\cdots+t+1\right)\right. \\
\left.\quad \quad+\left(t^{n-1}+t^{n-2}+\cdots+t+1\right)\right],
\end{array}
$$

Therefore, when $t>0$, and $t \neq \pm 1$, $a_{n+1}-a_{n}>0 \Rightarrow a_{n+1}>a_{n}$.