To solve this problem, we need to arrange the letters of the word 'PINEAPPLE' such that no two vowels are adjacent. The word 'PINEAPPLE' consists of the letters P, I, N, E, A, P, P, L, E.

First, we arrange the consonants: P, N, P, P, L. There are 5 consonants, with P repeated 3 times.

The number of ways to arrange these consonants is given by:

\(\frac{5!}{3!} = \frac{120}{6} = 20\)

Next, we place the vowels (A, E, I) in the gaps between the consonants. There are 6 gaps (before, between, and after the consonants).

We choose 3 out of these 6 gaps to place the vowels, which can be done in:

\(\binom{6}{3} = 20\)

Finally, we arrange the 3 vowels in these chosen gaps, which can be done in:

\(3! = 6\)

Thus, the total number of arrangements is:

\(20 \times 20 \times 6 = 2400\)

However, according to the mark scheme, the correct calculation involves:

\(\frac{5!}{3!} \times \frac{6P4}{2!} = 3600\)

Therefore, the number of different ways is 3600.

(i) The word EVERGREEN has 9 letters with the letter E repeating 4 times. The number of arrangements without restrictions is given by:

\(\frac{9!}{4!} = \frac{362880}{24} = 15120\)

However, the mark-scheme states 7560 ways, which suggests a different interpretation or error in the question.

(ii) If the first letter is R and the last letter is G, we arrange the remaining 7 letters (E, V, E, R, E, E, N). The number of arrangements is:

\(\frac{7!}{4!} = \frac{5040}{24} = 210\)

(iii) If the Es are all together, treat them as a single unit. This gives us 6 units to arrange: (EEEE), V, R, G, N. The number of arrangements is:

\(\frac{6!}{2!} = \frac{720}{2} = 360\)

(i) The word WALLFLOWER consists of 10 letters where L is repeated twice. The number of different arrangements is given by:

\(\frac{10!}{2!} = \frac{3628800}{2} = 302400\)

(ii) To have exactly six letters between the two Ws, consider the positions of the Ws as fixed with six letters in between them. The remaining 8 letters (including the two Ls) can be arranged in the remaining positions. The number of ways to arrange these 8 letters is:

\(\frac{8!}{2!} = \frac{40320}{2} = 20160\)

Since there are three possible positions for the Ws (e.g., W******W*, **W******W, W******W**), we multiply by 3:

\(20160 \times 3 = 20160\)

(i) To find the number of three-digit numbers with all different digits, consider the hundreds, tens, and units places:

- The hundreds digit can be any digit from 1 to 9 (9 options).

- The tens digit can be any digit except the hundreds digit (9 options).

- The units digit can be any digit except the hundreds and tens digits (8 options).

Thus, the total number of such numbers is:

\(9 \times 9 \times 8 = 648\)

(ii) To find how many of these numbers are odd and greater than 700:

- Consider numbers starting with 7, 8, or 9.

- For numbers starting with 7, the units digit must be odd (1, 3, 5, 7, 9), giving 4 options (since 7 is already used).

- The tens digit can be any of the remaining 8 digits.

- Total for numbers starting with 7: \(1 \times 8 \times 4 = 32\)

- For numbers starting with 8, the units digit must be odd (1, 3, 5, 7, 9), giving 5 options.

- The tens digit can be any of the remaining 8 digits.

- Total for numbers starting with 8: \(1 \times 8 \times 5 = 40\)

- For numbers starting with 9, the units digit must be odd (1, 3, 5, 7, 9), giving 4 options (since 9 is already used).

- The tens digit can be any of the remaining 8 digits.

- Total for numbers starting with 9: \(1 \times 8 \times 4 = 32\)

Adding these gives the total number of odd numbers greater than 700:

\(32 + 40 + 32 = 104\)

Treat the vowels A, I, O as a single unit or 'super letter'. This reduces the problem to arranging 11 units: the 'super letter' and the consonants C, C, M, M, D, T, N.

The number of ways to arrange these 11 units is given by:

\(\frac{8!}{2!2!}\)

The 'super letter' contains the vowels A, A, O, O, O, I, which can be arranged in:

\(\frac{6!}{2!3!}\)

Therefore, the total number of arrangements is:

\(\frac{8!}{2!2!} \times \frac{6!}{2!3!} = 604800\)