Answer: (a)(i) \(30240\); (a)(ii) \(17280\); (b)(i) \(35\); (b)(ii) \(51\).

We start with the main method. Count the total arrangements first, then subtract restricted cases or treat grouped letters as a single block.

For part (a)(i), the first and last positions must both be women.

There are \(3\) choices for the first position and then \(2\) choices for the last position.

The remaining \(7\) people can be arranged in \(7!\) ways.

So the number of arrangements is

\(3\cdot2\cdot7!=30240.\)

For part (a)(ii), keep the four children together as one block.

Then there are \(6\) objects to arrange: the child block and the other \(5\) people.

The objects can be arranged in \(6!\) ways, and the four children inside the block can be arranged in \(4!\) ways.

Therefore the number of arrangements is

\(6!\cdot4!=17280.\)

For part (b)(i), if no man is chosen, choose all \(4\) singers from the \(3\) women and \(4\) children.

There are \(7\) non-men, so the number of selections is

\(\binom74=35.\)

For part (b)(ii), at least \(2\) women means either exactly \(2\) women or exactly \(3\) women.

For exactly \(2\) women:

\(\binom32\binom62=3\cdot15=45.\)

For exactly \(3\) women:

\(\binom33\binom61=1\cdot6=6.\)

Hence the total is

\(45+6=51.\)

This completes the solution and gives the required result.