We have the digits 1, 3, 5 (odd) and 6, 6, 6, 8 (even). We treat the even digits as a block and the odd digits as another block. The blocks can be arranged in 2 ways: (odd, even) or (even, odd).

Within the odd block, the digits can be arranged in \(3!\) ways.

Within the even block, the digits can be arranged in \(\frac{4!}{3!}\) ways because the digit 6 is repeated three times.

Thus, the total number of arrangements is \(3! \times \frac{4!}{3!} \times 2 = 6 \times 4 \times 2 = 48\).

To find the number of even 7-digit numbers, we can use two methods:

Method 1: Calculate the total number of arrangements and subtract the number of odd-ending numbers.

The total number of arrangements is \(\frac{7!}{3!}\) because the digit 6 is repeated three times.

The number of odd-ending numbers is \(\frac{6!}{3!}\) because we fix an odd digit at the end and arrange the remaining 6 digits.

Thus, the number of even numbers is \(\frac{7!}{3!} - \frac{6!}{2!} = 840 - 360 = 480\).

Method 2: Count the numbers ending in 8 and 6 separately.

Numbers ending in 8: \(\frac{6!}{3!} = 120\).

Numbers ending in 6: \(\frac{6!}{2!} = 360\).

Total even numbers: \(120 + 360 = 480\).