(i) To expand \((1+y)^6\) up to the term in \(y^2\), we use the binomial theorem:

\((1+y)^6 = \binom{6}{0} + \binom{6}{1}y + \binom{6}{2}y^2 + \ldots\)

Calculating the coefficients, we have:

\(\binom{6}{0} = 1\)

\(\binom{6}{1} = 6\)

\(\binom{6}{2} = 15\)

Thus, the expansion is \(1 + 6y + 15y^2\).

(ii) For \((1 + (px - 2x^2))^6\), we need the coefficient of \(x^2\).

Using the binomial theorem, the relevant terms are:

\(\binom{6}{1}(px)(-2x^2)^0 = 6px\)

\(\binom{6}{2}(px)^0(-2x^2)^1 = 15(-2x^2) = -30x^2\)

The coefficient of \(x^2\) is given by:

\(\binom{6}{1}(px)(-2x^2)^1 + \binom{6}{2}(px)^0(-2x^2)^2\)

\(= 6p(-2x^2) + 15(-2x^2)^2\)

\(= 6p(-2)x^2 + 15(-2)^2x^2\)

\(= -12px^2 + 60x^2\)

Setting the coefficient equal to 48:

\(-12p + 60 = 48\)

\(-12p = 48 - 60\)

\(-12p = -12\)

\(p = 1\)

However, according to the mark scheme, \(p = 2\).