(i) Consider the vowels A, U, S, T, R, L, I, A as a single unit or 'super letter'. This gives us 5 units to arrange: (AAAIU), S, T, R, L. The number of ways to arrange these 5 units is given by:

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

Within the 'super letter', the vowels can be arranged in \(\frac{5!}{3!} = 20\) ways due to the repetition of A. Therefore, the total number of arrangements is:

\(5! \times \frac{5!}{3!} = 2400\)

(ii) The letter T is fixed in the central position. The end positions can be filled by the consonants R, S, L. The number of ways to choose and arrange these consonants is:

\(^3P_2 = 6\)

The remaining 6 letters (A, U, S, R, L, I) can be arranged in:

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

Thus, the total number of arrangements is:

\(120 \times 6 = 720\)