(20 (12 points) Given the function $f(x)=\frac{2 x+3}{3 x}$, the sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=1, a_{n+1}=$ $f\left(\frac{1}{a_{n}}\right), n \in \mathbf{N}^{*}$.
(1) Find the general term formula of the sequence $\left\{a_{n}\right\}$;
(2) Let $T_{n}=a_{1} a_{2}-a_{2} a_{3}+a_{3} a_{4}-a_{4} a_{5}+\cdots-a_{2 n} a_{2 n+1}$, find $T_{n}$.
20 (1) Since
$$
a_{n+1}=f\left(\frac{1}{a_{n}}\right)=\frac{\frac{2}{a_{n}}+3}{\frac{3}{a_{n}}}=\frac{2+3 a_{n}}{3}=a_{n}+\frac{2}{3} \text {, }
$$

thus $\left\{a_{n}\right\}$ is an arithmetic sequence with a common difference of $\frac{2}{3}$. Also, $a_{1}=1$, so
$$
a_{n}=\frac{2}{3} n+\frac{1}{3} .
$$
(2)
$$
\begin{aligned}
T_{n} & =a_{1} a_{2}-a_{2} a_{3}+a_{3} a_{4}-a_{4} a_{5}+\cdots-a_{2 n} a_{2 n+1} \\
& =a_{2}\left(a_{1}-a_{3}\right)+a_{4}\left(a_{3}-a_{5}\right)+\cdots+a_{2 n}\left(a_{2 n-1}-a_{2 n+1}\right) \\
& =-\frac{4}{3}\left(a_{2}+a_{4}+\cdots+a_{2 n}\right) \\
& =-\frac{4}{3} \cdot \frac{n\left(\frac{5}{3}+\frac{4 n}{3}+\frac{1}{3}\right)}{2} \\
& =-\frac{4}{9}\left(2 n^{2}+3 n\right)
\end{aligned}
$$