(a) Consider the even digits 2, 2, 4, 6, 6, 6 as a single block. This block can be arranged in \(\frac{6!}{2!3!}\) ways due to the repeated digits. The odd digits 1, 3, 3 can be arranged in \(\frac{3!}{2!}\) ways. Therefore, the total number of arrangements is \(\frac{6!}{2!3!} \times \frac{3!}{2!} = 720\) ways.

(b) If the first and last digits are both odd, we can have either 1 or 3 at the ends. Consider the cases:

• 1 at the start and 3 at the end: The remaining digits are 2, 2, 3, 4, 6, 6, 6, which can be arranged in \(\frac{7!}{3!2!} = 420\) ways.
• 3 at the start and 1 at the end: The remaining digits are 2, 2, 3, 4, 6, 6, 6, which can be arranged in \(\frac{7!}{3!2!} = 420\) ways.
• 3 at both ends: The remaining digits are 1, 2, 2, 4, 6, 6, 6, which can be arranged in \(\frac{7!}{3!2!} = 420\) ways.

Adding these, the total number of arrangements is \(420 + 420 + 420 = 1260\) ways.