16 Given the sequence $\left\{a_{n}\right\}$, the adjacent terms $a_{2 k-1}$ and $a_{2 k}$ are the roots of the equation
$$
x^{2}-\left(3 k+2^{k}\right) x+3 k \cdot 2^{k}=0
$$

(1) Find the sum of the first $2 n$ terms of the sequence $\left\{a_{n}\right\}$, $S_{2 n}$;
(2) Let $f(n)=\frac{1}{2}\left(\frac{|\sin n|}{\sin n}+3\right)$,
$T_{n}=\frac{(-1)^{f(2)}}{a_{1} a_{2}}+\frac{(-1)^{f(3)}}{a_{3} a_{4}}+\frac{(-1)^{f(4)}}{a_{5} a_{6}}+\cdots+\frac{(-1)^{f(n+1)}}{a_{2 n-1} a_{2 n}}$.
Prove: $\frac{1}{6} \leqslant T_{n} \leqslant \frac{5}{24}\left(n \in \mathbf{N}^{*}\right)$.
(16 (1) The equation $x^{2}-\left(3 k+2^{k}\right) x+3 k \cdot 2^{k}=0$ has two roots $x_{1}=3 k$, $x_{2}=2^{k}$, then
$$
\begin{aligned}
S_{2 n} & =a_{1}+a_{2}+\cdots+a_{2 n} \\
& =(3+6+\cdots+3 n)+\left(2+2^{2}+\cdots+2^{n}\right) \\
& =\frac{3 n^{2}+3 n}{2}+2^{n+1}-2 .
\end{aligned}
$$
(2)
$$
T_{n}=\frac{1}{a_{1} a_{2}}+\frac{1}{a_{3} a_{4}}-\frac{1}{a_{5} a_{6}}+\cdots+\frac{(-1)^{f(n+1)}}{a_{2 n-1} a_{2 n}},
$$

So $T_{1}=\frac{1}{a_{1} a_{2}}=\frac{1}{6}, T_{2}=\frac{1}{a_{1} a_{2}}+\frac{1}{a_{3} a_{4}}=\frac{5}{24}$.
When $n \geqslant 3$,
$$
\begin{aligned}
T_{n} & =\frac{1}{6}+\frac{1}{a_{3} a_{4}}-\frac{1}{a_{5} a_{6}}+\cdots+\frac{(-1)^{f(n+1)}}{a_{2 n-1} a_{2 n}} \\
& \geqslant \frac{1}{6}+\frac{1}{a_{3} a_{4}}-\left(\frac{1}{a_{5} a_{6}}+\cdots+\frac{1}{a_{2 n-1} a_{2 n}}\right) \\
& \geqslant \frac{1}{6}+\frac{1}{6 \cdot 2^{2}}-\frac{1}{6}\left(\frac{1}{2^{3}}+\cdots+\frac{1}{2^{n}}\right) \\
& =\frac{1}{6}+\frac{1}{6 \cdot 2^{n}}>\frac{1}{6}
\end{aligned}
$$

At the same time
$$
\begin{aligned}
T_{n} & =\frac{5}{24}-\frac{1}{a_{5} a_{6}}-\frac{1}{a_{7} a_{8}}+\cdots+\frac{(-1)^{f(n+1)}}{a_{2 n-1} a_{2 n}} \\
& \leqslant \frac{5}{24}-\frac{1}{a_{5} a_{6}}+\left(\frac{1}{a_{7} a_{8}}+\cdots+\frac{1}{a_{2 n-1} a_{2 n}}\right) \\
& \leqslant \frac{5}{24}-\frac{1}{9 \cdot 2^{3}}+\frac{1}{9}\left(\frac{1}{2^{1}}+\cdots+\frac{1}{2^{n}}\right) \\
& =\frac{5}{24}-\frac{1}{9 \cdot 2^{n}}<\frac{5}{24} .
\end{aligned}
$$

In summary, when $n \in \mathbf{N}^{*}$, $\frac{1}{6} \leqslant T_{n} \leqslant \frac{5}{24}$.