(i) Treat all the A's as a single unit. This gives us the units: {AAA, T, N, Z, N, I}. There are 6 units in total. The number of ways to arrange these units is given by:

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

The division by \(2!\) accounts for the repetition of the letter N.

(ii) We need to arrange the letters such that consonants and vowels alternate, starting with a consonant. The consonants are {T, N, Z, N} and the vowels are {A, A, I}. The number of ways to arrange the consonants is:

\(\frac{4!}{2!}\)

The number of ways to arrange the vowels is:

\(\frac{4!}{3!}\)

Thus, the total number of arrangements is:

\(\frac{4!}{2!} \times \frac{4!}{3!} = 48\)