(i) To find the first three terms of \((2-x)^6\), we use the binomial expansion formula:

\((2-x)^6 = \sum_{k=0}^{6} \binom{6}{k} (2)^{6-k} (-x)^k\).

The first three terms are:

\(\binom{6}{0} (2)^6 (-x)^0 = 64\),

\(\binom{6}{1} (2)^5 (-x)^1 = -192x\),

\(\binom{6}{2} (2)^4 (-x)^2 = 240x^2\).

Thus, the first three terms are \(64 - 192x + 240x^2\).

(ii) We need the coefficient of \(x^2\) in \((1 + 2x + ax^2)(2-x)^6\).

From part (i), the expansion of \((2-x)^6\) gives us the terms \(64 - 192x + 240x^2\).

Consider the terms contributing to \(x^2\):

1. \(1 \cdot 240x^2 = 240x^2\)

2. \(2x \cdot (-192x) = -384x^2\)

3. \(ax^2 \cdot 64 = 64ax^2\)

The total coefficient of \(x^2\) is \(240 - 384 + 64a\).

Set this equal to 48:

\(240 - 384 + 64a = 48\)

\(64a = 192\)

\(a = 3\)

(i) To find the first three terms of \((2 + 3x)^5\), we use the binomial expansion formula:

\((2 + 3x)^5 = \sum_{k=0}^{5} \binom{5}{k} (2)^{5-k} (3x)^k\).

The first three terms are:

\(\binom{5}{0} (2)^5 (3x)^0 = 32\)

\(\binom{5}{1} (2)^4 (3x)^1 = 5 \times 16 \times 3x = 240x\)

\(\binom{5}{2} (2)^3 (3x)^2 = 10 \times 8 \times 9x^2 = 720x^2\)

Thus, the first three terms are \(32 + 240x + 720x^2\).

(ii) To find \(a\) such that there is no \(x^2\) term in \((1 + ax)(2 + 3x)^5\), we consider the expansion:

\((1 + ax)(32 + 240x + 720x^2)\).

The \(x^2\) term comes from:

\(1 \times 720x^2 + ax \times 240x = 720 + 240ax^2\).

Setting the coefficient of \(x^2\) to zero:

\(720 + 240a = 0\)

\(a = -3\)

(i) To find the first 3 terms in the expansion of \((2 + x^2)^5\), we use the binomial theorem:

\((2 + x^2)^5 = \binom{5}{0} 2^5 + \binom{5}{1} 2^4 x^2 + \binom{5}{2} 2^3 (x^2)^2 + \ldots\)

Calculating each term:

First term: \(\binom{5}{0} 2^5 = 32\)

Second term: \(\binom{5}{1} 2^4 x^2 = 5 \times 16 x^2 = 80x^2\)

Third term: \(\binom{5}{2} 2^3 (x^2)^2 = 10 \times 8 x^4 = 80x^4\)

Thus, the first three terms are \(32 + 80x^2 + 80x^4\).

(ii) Now, expand \((1 + x^2)^2\):

\((1 + x^2)^2 = 1 + 2x^2 + x^4\)

To find the coefficient of \(x^4\) in \((1 + x^2)^2(2 + x^2)^5\), consider the product of terms that result in \(x^4\):

- \(1 \times 80x^4 = 80x^4\)

- \(2x^2 \times 80x^2 = 160x^4\)

- \(x^4 \times 32 = 32x^4\)

Adding these, the coefficient of \(x^4\) is \(80 + 160 + 32 = 272\).

The binomial expansion of \((2+ax)^n\) gives the first three terms as:

1st term: \(2^n\)

2nd term: \(n \cdot 2^{n-1} \cdot (ax)\)

3rd term: \(\frac{n(n-1)}{2} \cdot 2^{n-2} \cdot (ax)^2\)

Given: 1st term = 32, so \(2^n = 32\). Therefore, \(n = 5\).

2nd term: \(5 \cdot 2^4 \cdot ax = -40x\). Solving for \(a\), we get:

\(80a = -40\)

\(a = -\frac{1}{2}\)

3rd term: \(\frac{5 \cdot 4}{2} \cdot 2^3 \cdot (ax)^2 = bx^2\). Substituting \(a = -\frac{1}{2}\), we have:

\(10 \cdot 8 \cdot \left(-\frac{1}{2}\right)^2 = b\)

\(b = 20\)

(i) To find the first three terms in the expansion of \((2-x)^6\), we use the binomial theorem:

\((2-x)^6 = \sum_{r=0}^{6} \binom{6}{r} (2)^{6-r} (-x)^r\)

The first three terms are:

\(\binom{6}{0} (2)^6 (-x)^0 = 64\)

\(\binom{6}{1} (2)^5 (-x)^1 = -192x\)

\(\binom{6}{2} (2)^4 (-x)^2 = 240x^2\)

Thus, the first three terms are \(64 - 192x + 240x^2\).

(ii) For the expansion of \((1+kx)(2-x)^6\), we need the coefficient of \(x^2\) to be zero.

The expansion of \((2-x)^6\) gives the \(x^2\) term as \(240x^2\).

When multiplying by \(1+kx\), the \(x^2\) term comes from:

\(k \times (-192x) \times x = -192kx^2\)

Thus, the coefficient of \(x^2\) is \(240 - 192k\).

Setting this to zero gives:

\(240 - 192k = 0\)

\(192k = 240\)

\(k = \frac{240}{192} = \frac{5}{4} = 1.25\)