Answer: \(4368\); \(n=8\).

(a) The number must be odd and greater than \(70000\), so the first digit must be \(7\), \(8\) or \(9\), and the last digit must be odd.

If the first digit is \(8\), the last digit can be any of \(1,3,5,7,9\), giving \(5\) choices. The middle three digits can then be chosen and arranged in

\({}^8P_3=8\cdot7\cdot6=336\)

ways. This gives

\(5\cdot336=1680.\)

If the first digit is \(7\), the last digit can be \(1,3,5\) or \(9\), giving \(4\) choices. The middle three digits can again be arranged in \(336\) ways, giving \(1344\).

If the first digit is \(9\), similarly the last digit can be \(1,3,5\) or \(7\), giving another \(1344\).

Therefore the total number is

\(1680+1344+1344=4368.\)

(b) The statement gives

\({}^nC_3=2{}^nC_2.\)

Using the formula for combinations,

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

Canceling \(n!\),

\(\frac1{6(n-3)!}=\frac1{(n-2)!}.\)

Since \((n-2)!=(n-2)(n-3)!\),

\(n-2=6.\)

Hence

\(n=8.\)