Answer: \(660\); \(n=11\).

(a) The number must be even, so its last digit must be one of \(0,4,6\).

If the first digit is \(6\), the last digit can be \(0\) or \(4\). The middle three digits can then be arranged in \(5P3\) ways. This gives

\(2\cdot {}^5P_3=2\cdot 60=120.\)

If the first digit is \(5\), \(7\) or \(9\), there are \(3\) choices for the first digit and \(3\) choices for the last digit. The middle three digits can again be arranged in \(5P3\) ways. This gives

\(3\cdot3\cdot{}^5P_3=9\cdot60=540.\)

Therefore the total number is

\(120+540=660.\)

(b) Use the formula for combinations:

\({}^nC_4=6{}^nC_2.\)

So

\(\frac{n!}{4!(n-4)!}=6\cdot\frac{n!}{2!(n-2)!}.\)

Canceling \(n!\) and simplifying gives

\(\frac{(n-2)(n-3)}{24}=3.\)

Hence

\((n-2)(n-3)=72.\)

So

\(n^2-5n-66=0.\)

Factorising,

\((n-11)(n+6)=0.\)

Since \(n\) is positive, \(n=11\).