(i) To find the number of possible arrangements of the 12 houses, we use the formula for permutations of multiset:

\(\frac{12!}{2!2!3!4!}\)

Calculating this gives:

\(\frac{479001600}{4 \times 2 \times 6 \times 24} = 831600\)

(ii) For the second part, we need to arrange the houses in two groups of 6. In the first group, we have all houses in styles A and D, which are 2 and 4 respectively. The number of arrangements for the first group is:

\(\frac{6!}{4!2!}\)

In the second group, we have all houses in styles B, C, and E, which are 2, 3, and 1 respectively. The number of arrangements for the second group is:

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

The total number of arrangements is the product of the arrangements of the two groups:

\(\frac{6!}{4!2!} \times \frac{6!}{2!3!} = 900\)