(a) To arrange the letters such that no consonant is next to another consonant, we first arrange the vowels A, O, E, A. There are 4 vowels, and they can be arranged in \(\frac{4!}{2!}\) ways due to the repetition of A. This gives \(\frac{4!}{2!} = 12\) ways.

Next, we place the 5 consonants (D, M, N, R, D) in the gaps between and around the vowels. There are 5 gaps (before, between, and after the vowels), and we need to choose 4 of these gaps to place the consonants. The consonants can be arranged in \(\frac{5!}{2!}\) ways due to the repetition of D. This gives \(\frac{5!}{2!} = 60\) ways.

The total number of arrangements is \(12 \times 60 = 720\).

(b) To arrange the letters with an A at each end and the Ds not together, we first fix the As at each end: A _ _ _ _ _ _ _ A. This leaves 7 positions for the remaining letters (D, N, D, R, O, M, E).

First, calculate the total arrangements with A at each end: \(\frac{7!}{2!} = 2520\) ways, due to the repetition of D.

Next, calculate the arrangements where the Ds are together. Treat the two Ds as a single unit, so we have (DD, N, R, O, M, E) to arrange, which is \(6! = 720\) ways.

The number of arrangements where the Ds are not together is \(2520 - 720 = 1800\).