(a) The word CELESTIAL has 9 letters with 2 Es. The number of different arrangements is given by:

\(\frac{9!}{2!} = 90,720\)

(b) Fixing C as the first letter, T as the fifth letter, and E as the last letter, we arrange the remaining 6 letters (E, L, S, I, A, L). The number of arrangements is:

\(\frac{6!}{2!} = 360\)

(c) First, calculate the number of arrangements where the two Es are together. Treat the two Es as a single unit, reducing the problem to arranging 8 units (EE, C, L, S, T, I, A, L):

\(\frac{8!}{2!} = 20,160\)

The number of arrangements where the Es are not together is:

\(90,720 - 20,160 = 70,560\)

The probability that the Es are not together is:

\(\frac{70,560}{90,720} = \frac{7}{9} \approx 0.778\)