【Analysis】(1) It is easy to know that a rope of length 2 can only be divided into two segments of length 1, meaning the last operation on the rope must not be a good operation. A rope of length 2009 can be divided into 2009 segments of length 1 after exactly 2008 operations, so the number of good operations is no more than
$2008-1=2007$.
Therefore, the maximum number of good operations is 2007.
The following is such an operation: each time, cut a segment of length 1, then the first 2007 operations are all good operations.

Generalize the problem: For a rope of length $n$, let $f(n)$ be the minimum number of good operations.

By conducting mathematical experiments with simple positive integers $n$ and summarizing the experience, the following two lemmas can be discovered.
Lemma 1: $f(n)=0 \Leftrightarrow n=2^{k}$.
Lemma 2: Suppose $n=2^{k_{1}}+2^{k_{2}}+\cdots+2^{k_{1}}\left(k_{1}>k_{2}\right.$ $\left.>\cdots>k_{l} \geqslant 0\right)$. Let $l=S_{2}(n) \triangleq$ the sum of the digits of $n$ in binary. Then $f(n)=S_{2}(n)-1$.

Proof of Lemma 2: Since after $l-1$ good operations, we can obtain $2^{k_{1}}, 2^{k_{2}}, \cdots, 2^{k_{1}}$, then
$f(n) \leqslant l-1$.
We will prove by mathematical induction:
$f(n) \geqslant S_{2}(n)-1$.
When $n=1$, it is Lemma 1, and the conclusion is obviously true.
Assume that for $n \leqslant k$, $f(n) \geqslant S_{2}(n)-1$.
When $n=k+1$, by the definition of $f$, there exists an operation process containing $f(k+1)$ good operations and several non-good operations, which cuts a rope of length $k+1$ into $k+1$ segments of length 1. If the first operation in the above process is a good operation, cutting a rope of length $k+1$ into segments of lengths $a$ and $b$, i.e., $k+1=a+b, a \neq b$, then
$$\begin{array}{l} f(k+1)=f(a)+f(b)+1 \\ \geqslant\left(S_{2}(a)-1\right)+\left(S_{2}(b)-1\right)+1 \\ =S_{2}(a)+S_{2}(b)-1 \\ \geqslant S_{2}(a+b)-1=S_{2}(k+1)-1, \end{array}$$

The equality holds if and only if there is no carry in the binary addition of $a+b$.

If the first operation in the above process is not a good operation, i.e., $k+1=2a$, then
$$f(k+1)=2 f(a) \geqslant 2\left(S_{2}(a)-1\right) .$$

Since $S_{2}(2a) \leqslant S_{2}(a)+S_{2}(a)-1$, the equality holds if and only if $S_{2}(a)=1$, i.e.,
$$2 S_{2}(a)-2 \geqslant S_{2}(2a)-1 .$$

Using this result, we get
$$\begin{array}{l} f(k+1) \geqslant 2\left(S_{2}(a)-1\right) \\ \geqslant S_{2}(2a)-1=S_{2}(k+1)-1, \end{array}$$

The conclusion holds, and the equality holds if and only if $S_{2}(a)=1$.
In summary, Lemma 2 is proved.
Therefore, using the binary representation of 2009, we get
$$f(2009)=S_{2}(2009)-1=7 \text {. }$$
(2) If the binary representation of $n$ is
$$n=2^{k_{1}}+2^{k_{2}}+\cdots+2^{k_{i}} \text {, }$$

we can set the set $M=\left\{k_{1}, k_{2}, \cdots, k_{l}\right\}$.
If $l \geqslant 2$, then in any operation process where the number of good operations reaches the minimum, the first operation is a good operation, i.e., there exists a non-empty proper subset $L$ of $M$ such that the rope of length $n$ is cut into segments of lengths $a$ and $b$, where $a=\sum_{k \in L} 2^{k}, b=\sum_{k \in M-L} 2^{k}$. Performing $l-1$ good operations in succession produces $l$ different lengths $2^{k_{1}}, 2^{k_{2}}, \cdots, 2^{k_{i}}$.

Then, the non-good operations produce lengths $2^{k_{1}-1}, 2^{k_{1}-2}, \cdots, 2,2^{0}$, and so on. Therefore, the number of different lengths recorded is $l-1+k_{1}$.

In solving this problem, it is found through mathematical experiments that the minimum number of good operations $f(n)$ is related to the binary representation of the positive integer $n$, and a basic result about the sum of digits is used:
$$S_{p}(a+b)=S_{p}(a)+S_{p}(b)-k(p-1),$$

where $k$ is the number of carries in the addition of $a+b$ in base $p$.