(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 \(400\).

(i) To find the term in \(x^2\) in the expansion of \((1 - \frac{3}{2}x)^6\), use the binomial theorem:

\(^6C_2 \left( -\frac{3}{2}x \right)^2 = 15 \times \left( \frac{9}{4}x^2 \right) = 135x^2.\)

For the term in \(x^3\):

\(^6C_3 \left( -\frac{3}{2}x \right)^3 = 20 \times \left( -\frac{27}{8}x^3 \right) = -540x^3.\)

(ii) For the expansion of \((k + 2x)(1 - \frac{3}{2}x)^6\), the term in \(x^3\) is:

\(2 \times 135x^2 + k \times (-540x^3) = 0.\)

\(\frac{270x^3}{4} - \frac{135kx^3}{2} = 0.\)

Solving for \(k\):

\(270 - 270k = 0\)

\(k = 1.\)

The general term in the binomial expansion of \((1 + ax)^6\) is given by:

\(T_k = \binom{6}{k} (1)^{6-k} (ax)^k = \binom{6}{k} a^k x^k\)

For the term in \(x\), we have \(k = 1\):

\(T_1 = \binom{6}{1} a^1 x^1 = 6ax\)

We know the coefficient of \(x\) is \(-30\), so:

\(6a = -30\)

Solving for \(a\):

\(a = -5\)

Now, find the coefficient of \(x^3\) (\(k = 3\)):

\(T_3 = \binom{6}{3} a^3 x^3\)

\(\binom{6}{3} = \frac{6 \times 5 \times 4}{3!} = 20\)

Substitute \(a = -5\):

\(T_3 = 20(-5)^3 x^3\)

\(T_3 = 20(-125)x^3\)

\(T_3 = -2500x^3\)

Thus, the coefficient of \(x^3\) is \(-2500\).

(i) The binomial expansion of \((k + x)^8\) is given by:

\(\sum_{r=0}^{8} \binom{8}{r} k^{8-r} x^r\)

The first four terms are:

\(\binom{8}{0} k^8 x^0 = k^8\)

\(\binom{8}{1} k^7 x^1 = 8k^7x\)

\(\binom{8}{2} k^6 x^2 = 28k^6x^2\)

\(\binom{8}{3} k^5 x^3 = 56k^5x^3\)

Thus, the first four terms are \(k^8 + 8k^7x + 28k^6x^2 + 56k^5x^3\).

(ii) Given that the coefficients of \(x^2\) and \(x^3\) are equal, we have:

\(28k^6 = 56k^5\)

Dividing both sides by \(28k^5\), we get:

\(k = 2\)