(a) Consider the three Es as a single unit, reducing the problem to arranging 7 units: \(\{EEE, D, D, L, I, V, R\}\). The total arrangements of these 7 units is \(\frac{7!}{2!}\) due to the repetition of the two Ds. This gives \(\frac{5040}{2} = 2520\).

Now, subtract the arrangements where the two Ds are together. Treat the two Ds as a single unit, reducing the problem to arranging 6 units: \(\{EEE, DD, L, I, V, R\}\). The total arrangements of these 6 units is \(6! = 720\).

Thus, the number of arrangements where the three Es are together and the two Ds are not next to each other is \(2520 - 720 = 1800\).

(b) The total number of arrangements of the 9 letters is \(\frac{9!}{3!2!} = 362880\).

To find the number of arrangements with exactly 4 letters between the two Ds, consider the positions of the Ds. If the first D is in position 1, the second D can be in position 6. The letters between them can be any 4 of the remaining 7 letters. The number of ways to choose 4 letters from 7 is \(\binom{7}{4} = 35\).

For each choice of 4 letters, there are \(4!\) ways to arrange them. Thus, the number of arrangements for this scenario is \(35 \times 24 = 840\).

Since there are 5 possible starting positions for the first D (1 to 5), the total number of favorable arrangements is \(5 \times 840 = 4200\).

The probability is then \(\frac{4200}{362880} = \frac{1}{9}\).