(i) Consider the odd digits (3, 7, 7, 7) as one block and the even digits (2, 2, 8) as another block. The arrangement of these two blocks can be done in 2! ways.

Within the odd block, the digits can be arranged as \(\frac{4!}{3!}\) because there are three 7s.

Within the even block, the digits can be arranged as \(\frac{3!}{2!}\) because there are two 2s.

Therefore, the total number of arrangements is \(2! \times \frac{4!}{3!} \times \frac{3!}{2!} = 2 \times 4 \times 3 = 24\).

(ii) First, calculate the total number of arrangements of the digits without any restriction: \(\frac{7!}{2!3!} = 420\).

Next, calculate the number of arrangements where the two 2s are together. Treat the two 2s as a single unit, so we have 6 units to arrange: \(\frac{6!}{3!} = 120\).

The number of arrangements where the 2s are not together is the total arrangements minus the arrangements where the 2s are together: \(420 - 120 = 300\).