3.3. ** Given a finite sequence $\left\{a_{n}\right\}$ with $2 k$ terms (integer $k \geqslant 2$), the first term $a_{1}=2$. Let the sum of the first $n$ terms be $S_{n}$, and $a_{n+1}=(a-1) S_{n}+2(n=1,2, \cdots, 2 k-1)$, where the constant $a>1$.
(1) Prove that the sequence $\left\{a_{n}\right\}$ is a geometric sequence;
(2) If $a=2^{\frac{2}{2 k-1}}$, the sequence $\left\{b_{n}\right\}$ satisfies $b_{n}=\frac{1}{n} \log _{2}\left(a_{1} a_{2} \cdots a_{n}\right)(n=1,2, \cdots, 2 k)$, find the general term formula for the sequence $\left\{b_{n}\right\}$;
(3) If the sequence $\left\{b_{n}\right\}$ in (2) satisfies the inequality $\left|b_{1}-\frac{3}{2}\right|+\left|b_{2}-\frac{3}{2}\right|+\cdots+\mid b_{2 k-1}-$ $\left.\frac{3}{2}|+| b_{2 k}-\frac{3}{2} \right\rvert\, \leqslant 4$, find the value of $k$.
(1) When $n=1$, $a_{2}=2 a$, then $\frac{a_{2}}{a_{1}}=a$;
When $2 \leqslant n \leqslant 2 k-1$, $a_{n+1}=(a-1) S_{n}+2, a_{n}=(a-1) S_{n-1}+2, a_{n+1}-$ $a_{n}=(a-1) a_{n}$, so $\frac{a_{n+1}}{a_{n}}=a$, which means the sequence $\left\{a_{n}\right\}$ is a geometric sequence.
(2) From (1), we get $a_{n}=2 a^{n-1}$, so $a_{1} a_{2} \cdots \cdots \cdot a_{n}=2^{n} a^{1+2+\cdots+(n-1)}=2^{n} a^{\frac{(n-1) n}{2}}=$ $2^{n+\frac{n(n-1)}{2 n-1}}$,
$$
b_{n}=\frac{1}{n}\left[n+\frac{n(n-1)}{2 k-1}\right]=\frac{n-1}{2 k-1}+1 . \quad(n=1,2, \cdots, 2 k)
$$
(3) Let $b_{n} \leqslant \frac{3}{2}$, solving gives $n \leqslant k+\frac{1}{2}$. Since $n$ is a positive integer, when $n \leqslant k$, $b_{n}\frac{3}{2}$.
$$
\begin{aligned}
\text { Original expression }= & \left(\frac{3}{2}-b_{1}\right)+\left(\frac{3}{2}-b_{2}\right)+\cdots+\left(\frac{3}{2}-b_{k}\right) \\
& +\left(b_{k+1}-\frac{3}{2}\right)+\cdots+\left(b_{2 k}-\frac{3}{2}\right) \\
= & \left(b_{k+1}+\cdots+b_{2 k}\right)-\left(b_{1}+\cdots+b_{k}\right) \\
= & {\left[\frac{\frac{1}{2}(k+2 k-1) k}{2 k-1}+k\right]-\left[\frac{\frac{1}{2}(0+k-1) k}{2 k-1}+k\right]=\frac{k^{2}}{2 k-1} . }
\end{aligned}
$$

From $\frac{k^{2}}{2 k-1} \leqslant 4$, we get $k^{2}-8 k+4 \leqslant 0,4-2 \sqrt{3} \leqslant k \leqslant 4+2 \sqrt{3}$.
Since $k \geqslant 2$, when $k=2,3,4,5,6,7$, the original inequality holds.