(1) Solution 1: Let the number of large trucks be $x$ and the number of small trucks be $y$. According to the problem, we have

$$\begin{align*} x + y &= 20 \\ 15x + 10y &= 240 \end{align*}$$

Solving these equations, we get

$$\begin{align*} x &= 8 \\ y &= 12 \end{align*}$$

Therefore, 8 large trucks and 12 small trucks are used.

Solution 2: Let the number of large trucks be $x$ and the number of small trucks be $(20-x)$. According to the problem, we have

$$15x + 10(20-x) = 240$$

Solving for $x$, we get $x=8$.

Therefore, $20-x = 20-8 = 12$ (trucks).

Therefore, 8 large trucks and 12 small trucks are used.

(2) Let the total freight charges be $W$ yuan. Let the number of large trucks going to location A be $a$, and the number of small trucks going to location A be $(10-a)$. The number of large trucks going to location B is $(8-a)$, and the number of small trucks going to location B is $12-(10-a) = (a+2)$. Thus,

$$W = 630a + 420(10-a) + 750(8-a) + 550(a+2)$$

Simplifying, we get:

$$W = 10a + 11300 \quad (0 \leq a \leq 8, \text{ where } a \text{ is an integer})$$

Since $15a + 10(10-a) \geq 115$, we have $a \geq 3$.

Since $W$ increases with $a$, the minimum $W$ occurs when $a=3$.

When $a=3$, $W = 10 \times 3 + 11300 = 11330$.

Therefore, 3 large trucks and 7 small trucks should be arranged to go to location A, and 5 large trucks and 5 small trucks to location B, with the minimum total freight charges being $\boxed{11330 \text{ yuan}}$.