(i) To expand \((2-y)^5\), use the binomial theorem:

\((2-y)^5 = \sum_{k=0}^{5} \binom{5}{k} (2)^{5-k} (-y)^k\).

The first three terms are:

\(\binom{5}{0} (2)^5 (-y)^0 = 32\)

\(\binom{5}{1} (2)^4 (-y)^1 = -80y\)

\(\binom{5}{2} (2)^3 (-y)^2 = 80y^2\)

Thus, the first three terms are \(32 - 80y + 80y^2\).

(ii) Substitute \(y = 2x + x^2\) into the expansion:

\((2-(2x-x^2))^5 = (2-(2x+x^2))^5 = (2-y)^5\).

Using the result from part (i), the terms involving \(x^2\) are:

\(-80(2x) + 80(x^2)\).

Calculate the coefficient of \(x^2\):

\(-80 \cdot 2x + 80x^2 = -160x + 80x^2\).

Combine terms to find the coefficient of \(x^2\):

\(80 \cdot 4 = 320\).

Thus, the coefficient of \(x^2\) is \(400\).