(i) To find the first three terms in the expansion of \((1 + x)^5\), we use the binomial theorem:

\((1 + x)^5 = \sum_{k=0}^{5} \binom{5}{k} x^k\)

The first three terms are:

\(\binom{5}{0} x^0 = 1\)

\(\binom{5}{1} x^1 = 5x\)

\(\binom{5}{2} x^2 = 10x^2\)

Thus, the first three terms are \(1 + 5x + 10x^2\).

(ii) For the expansion of \((1 + (px + x^2))^5\), we replace \(x\) by \((px + x^2)\) in the expansion of \((1 + x)^5\):

\((1 + (px + x^2))^5 = 1 + 5(px + x^2) + 10(px + x^2)^2\)

We need the coefficient of \(x^2\):

From \(5(px + x^2)\), the \(x^2\) term is \(5x^2\).

From \(10(px + x^2)^2\), the \(x^2\) term is \(10p^2x^2\).

Thus, the coefficient of \(x^2\) is \(5 + 10p^2\).

We are given that this coefficient is 95:

\(5 + 10p^2 = 95\)

\(10p^2 = 90\)

\(p^2 = 9\)

\(p = 3\) (since \(p\) is positive)