(a) First, calculate the total number of arrangements of the letters in CASABLANCA. The word has 10 letters with the following repetitions: A appears 3 times, C appears 2 times. The total arrangements are given by:

\(\frac{10!}{2!4!} = 75600\)

Next, calculate the number of arrangements where the two Cs are together. Treat the two Cs as a single unit, reducing the problem to arranging 9 units (CC, A, S, A, B, L, A, N, A). The arrangements are:

\(\frac{9!}{4!} = 15120\)

Thus, the number of arrangements where the Cs are not together is:

\(75600 - 15120 = 60480\)

(b) Fix A at the beginning and A at the end, leaving 8 positions to fill. The sequence is A _ _ _ C _ _ _ C _ A. The three letters between the Cs can be chosen from the remaining 6 letters (S, A, B, L, N, A). The number of ways to choose and arrange these 3 letters is:

\(\frac{6!}{2!} \times 4 = 1440\)