(i) The number 1223678 has 7 digits, with the digit '2' repeated twice. The number of different permutations is given by:

\(\frac{7!}{2!} = \frac{5040}{2} = 2520\)

(ii) Consider the digits 1, 3, 7 as a single block. This block, along with the other 4 digits (2, 2, 6, 8), forms 5 blocks in total. The number of permutations of these 5 blocks is:

\(\frac{5!}{2!} = \frac{120}{2} = 60\)

Within the block of 1, 3, 7, the digits can be arranged in \(3!\) ways. Therefore, the total number of permutations is:

\(60 \times 6 = 360\)

(iii) For the number to be even, it must end in 2, 6, or 8. Calculate the permutations for each case:

Ending in 2: Consider the remaining digits 1, 2, 3, 6, 7. The number of permutations is:

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

Ending in 6 or 8: The number of permutations for each is:

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

Summing these gives:

\(360 + 360 + 360 = 1440\)