a) Let the length of the strip be $n$. Denote by $g(n)$ and $f(n)$ respectively the number of $1$'s in the binary representation of $n$ and the minimum possible number of bad cuts. Let $n = 2^{k_1} + 2^{k_2} + \cdots + 2^{k_l}$. Consider the following sequence of cuts: first cut a strip of length $2^{k_1}$, next cut a strip of length $2^{k_2}$ and so on. After the last cut we obtain two strips of lengths $2^{k_{l-1}}$ and $2^{k_l}$. Note that a strip whose length is a power of $2$ can be cut into strips of length $1$ without bad cuts. Therefore the number of bad cuts equals $l-1$, i.e.
$$(1) \qquad f(n) \le g(n) - 1.$$
We prove by induction on $n$ that $f(n) \ge g(n) - 1$. For $n=1$ we have $f(1) = 0$ and $g(1) = 1$, i.e. the statement is true. Suppose it is true for all $n \le k$, where $k$ is a positive integer and let $n = k+1$.

1. Suppose the first cut is bad and it leaves two strips of lengths $a$ and $b$. Then $a+b = k+1$ and $f(k+1) = 1 + f(a) + f(b)$. If the binary representations of $a$ and $b$ have no common digit $1$ then $g(k+1) = g(a) + g(b)$ and therefore
$$f(k+1) = 1 + f(a) + f(b) \ge 1 + g(a) - 1 + g(b) - 1 = g(k+1) - 1.$$
If the binary representations of $a$ and $b$ have at least one common digit $1$ then $g(k+1) \le g(a) + g(b) - 1$ and therefore
$$f(k+1) = 1 + f(a) + f(b) \ge 1 + g(a) - 1 + g(b) - 1 \ge g(k+1) > g(k+1) - 1.$$

2. Suppose the first cut is not bad, i.e. the strip is cut into two parts each of length $a$. Then $k+1 = 2a$ and $g(k+1) = g(a)$. If $g(k+1) = 1$ then $f(k+1) = 0$ and the statement is true. Otherwise
$$f(k+1) = f(a) + f(b) = 2f(a) \ge 2g(a) - 2 = 2g(k+1) - 2 > g(k+1) - 1.$$
Thus, when $g(k+1) > 1$ we have $f(k+1) > g(k+1) - 1$.

This proves the induction hypothesis giving
$$(2) \qquad f(n) \geq g(n) - 1.$$
It follows from (1) and (2) that $f(n) = g(n) - 1$.

a) Since the binary representation of $2007$ is $1111010111$, i.e. $g(2007) = 9$, we obtain $f(2007) = 8$.

b) It follows from the above arguments that if the number of bad cuts is $f(n) = g(n) - 1$ then every bad cut leaves two parts of lengths $a$ and $b$ such that the binary representations of $a$ and $b$ have no common digit $1$. Moreover the good cuts are done only over strips whose lengths are powers of $2$. It is clear that rearranging the cuts we may assume that the first cuts are bad. Their number equals $g(n) - 1$ and each bad cut gives two new numbers on the table. Therefore after all bad cuts we have $2g(n) - 2$ distinct numbers. The powers of $2$ that appear are all powers up to the highest power in the binary representation of $n$.
Thus, the number of distinct numbers on the table equals $2g(n) - 2 + k + 1 = 2g(n) + k - 1$, where $k$ is the highest power of $2$ in the binary representation of $n$.