(i) The formula for the nth term of an arithmetic progression is given by:

\(a_n = a + (n-1)d\)

For the ninth term, \(a + 8d = 22\).

The sum of the first 4 terms is given by:

\(S_4 = \frac{4}{2}(2a + 3d) = 49\)

\(2a + 3d = 24.5\)

We have the system of equations:

1. \(a + 8d = 22\)

2. \(2a + 3d = 24.5\)

Solving these equations simultaneously, we find:

\(d = 1.5\)

\(a = 10\)

(ii) The nth term is given by:

\(a + (n-1)d = 46\)

Substituting \(a = 10\) and \(d = 1.5\):

\(10 + (n-1) \times 1.5 = 46\)

\(1.5(n-1) = 36\)

\(n-1 = 24\)

\(n = 25\)