(i) Let the common difference be \(d\). The sum of the first ten terms \(S_{10}\) is given by:

\(S_{10} = \frac{10}{2} (2a + 9d) = 5(2a + 9d)\)

The sum of the next five terms \(S_{11 \text{ to } 15}\) is:

\(S_{11 \text{ to } 15} = \frac{5}{2} [(a + 10d) + (a + 14d)] = 5(a + 12d)\)

Given \(S_{10} = S_{11 \text{ to } 15}\), we have:

\(5(2a + 9d) = 5(a + 12d)\)

\(2a + 9d = a + 12d\)

\(a = 3d\)

Thus, \(d = \frac{a}{3}\).

(ii) The tenth term \(a + 9d\) is 36 more than the fourth term \(a + 3d\):

\(a + 9d = (a + 3d) + 36\)

\(6d = 36\)

\(d = 6\)

Since \(d = \frac{a}{3}\), we have:

\(\frac{a}{3} = 6\)

\(a = 18\)