8. (2001 Shanghai Competition Problem) A publishing company prices a best-selling book as follows:
$$
C(n)=\left\{\begin{array}{l}
12 n, \text { if } 1 \leqslant n \leqslant 24 ; \\
11 n, \text { if } 25 \leqslant n \leqslant 48 ; \\
10 n, \text { if } n \geqslant 49 .
\end{array}\right.
$$

Here $n$ represents the number of books ordered, and $C(n)$ is the amount of money (in yuan) paid for ordering $n$ books.
(1) How many $n$ are there such that buying more than $n$ books costs less than buying exactly $n$ books?
(2) If the cost of one book is 5 yuan, and two people come to buy books, each buying at least 1 book, and together they buy 60 books, what is the minimum and maximum profit the publishing company can make?
8. (1) Given $C(25)=275, C(24)=288, C(23)=276, C(22)=264$.

We have $C(25)<C(23)<C(24)$.
Given $C(49)=490, C(48)=528, C(47)=517, C(46)=506, C(45)=495, C(44)=484$,
We have $C(49)<C(45)<C(46)<C(47)<C(48)$.
Therefore, there are 6 values of $n$ (i.e., $23, 24, 45, 46, 47, 48$) where buying more than $n$ books costs less than buying exactly $n$ books.
(2) Let the two individuals buy $a$ and $b$ books respectively, paying a total of $s$ yuan. Without loss of generality, assume $a \leqslant b$. From $a+b=60$, we know $1 \leqslant a \leqslant 30$.
(i) When $1 \leqslant a \leqslant 11$, $49 \leqslant b \leqslant 59$,
$$
s=12 a+10 b=10(a+b)+2 a=600+2 a, 602 \leqslant s \leqslant 622 \text{. }
$$
(ii) When $12 \leqslant a \leqslant 24$, $36 \leqslant b \leqslant 48$,
$$
s=12 a+11 b=660+a, 672 \leqslant s \leqslant 684 \text{. }
$$
(iii) When $25 \leqslant a \leqslant 30$, $30 \leqslant b \leqslant 35$,
$$
s=11 a+11 b=660 \text{. }
$$

Therefore, the publishing company earns at least $602-60 \times 5=302$ (yuan) and at most $684-60 \times 5=384$ (yuan).