Answer: (a) \(28560\); (b) \(n=207\).

We start with the main method. Use combinations when order does not matter, and split the count into valid cases if there are lower bounds.

For a number to be divisible by \(5\), its final digit must be \(0\) or \(5\).

First consider numbers ending in \(0\).

The final digit is fixed as \(0\), and the first five positions are filled from the remaining \(9\) digits.

So the number of arrangements is

\({}^{9}P_5=9\cdot8\cdot7\cdot6\cdot5=15120.\)

Now consider numbers ending in \(5\).

The final digit is fixed as \(5\). The first digit cannot be \(0\), so it has \(8\) choices.

After choosing the first digit, the middle four positions can be filled from \(8\) remaining digits.

So the number of arrangements is

\(8\cdot{}^{8}P_4=8\cdot8\cdot7\cdot6\cdot5=13440.\)

Therefore the total number of 6-digit numbers divisible by \(5\) is

\(15120+13440=28560.\)

For part (b), the condition is

\(\binom{n+1}{13}=16\binom n{12}.\)

Now

\(\binom{n+1}{13}=\frac{(n+1)!}{13!(n-12)!}\)

and

\(\binom n{12}=\frac{n!}{12!(n-12)!}.\)

So

\(\frac{\binom{n+1}{13}}{\binom n{12}}=\frac{n+1}{13}.\)

Since \(\binom{n+1}{13}=16\binom n{12}\),

\(\frac{n+1}{13}=16.\)

Thus

\(n+1=208\), and hence

\(n=207.\)

This completes the solution and gives the required result.