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

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

The first three terms are:

\(\binom{5}{0} 2^5 u^0 = 32\),

\(\binom{5}{1} 2^4 u^1 = 80u\),

\(\binom{5}{2} 2^3 u^2 = 80u^2\).

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

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

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

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

From \(80(x + x^2)\), the coefficient of \(x^2\) is \(80\).

From \(80(x + x^2)^2\), expand \((x + x^2)^2 = x^2 + 2x^3 + x^4\), the coefficient of \(x^2\) is \(80\).

Adding these gives the total coefficient of \(x^2\) as \(80 + 80 = 160\).

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

\(\binom{6}{3} \left( \frac{2}{x} \right)^3 (-3x)^3\)

Calculating each part:

\(\binom{6}{3} = 20\)

\(\left( \frac{2}{x} \right)^3 = \frac{8}{x^3}\)

\((-3x)^3 = -27x^3\)

Multiplying these gives:

\(20 \times \frac{8}{x^3} \times (-27x^3) = -4320\)

Thus, the term independent of \(x\) is \(-4320\).

(ii) To find the value of \(a\) for which there is no term independent of \(x\) in the expansion of \(\left( 1 + ax^2 \right) \left( \frac{2}{x} - 3x \right)^6\), we consider the critical term:

\(\binom{6}{2} \left( \frac{2}{x} \right)^4 (-3x)^2\)

Calculating each part:

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

\(\left( \frac{2}{x} \right)^4 = \frac{16}{x^4}\)

\((-3x)^2 = 9x^2\)

Multiplying these gives:

\(15 \times \frac{16}{x^4} \times 9x^2 = \frac{2160}{x^2}\)

For no term independent of \(x\), the coefficient of \(x^0\) must be zero:

\(15a \times 16 \times 9 - 4320 = 0\)

Solving for \(a\):

\(2160a = 4320\)

\(a = 2\)

(i) The general term in the expansion of \(\left( x - \frac{2}{x} \right)^6\) is given by \(T_r = \binom{6}{r} x^{6-r} \left( -\frac{2}{x} \right)^r\).

For the term to be independent of \(x\), the powers of \(x\) must cancel out, i.e., \(6 - r - r = 0\), which gives \(r = 3\).

Substituting \(r = 3\) into the general term, we get:

\(T_3 = \binom{6}{3} x^{6-3} \left( -\frac{2}{x} \right)^3 = 20 \cdot x^3 \cdot \left( -\frac{8}{x^3} \right) = -160\).

Thus, the term independent of \(x\) is \(-160\).

(ii) The term independent of \(x\) in the product \(\left( 2 + \frac{3}{x^2} \right) \left( x - \frac{2}{x} \right)^6\) is found by considering the contributions from both terms.

First, find the term in \(\left( x - \frac{2}{x} \right)^6\) that results in \(x^2\):

\(T_r = \binom{6}{r} x^{6-r} \left( -\frac{2}{x} \right)^r\).

For \(x^2\), \(6 - r - r = 2\), which gives \(r = 2\).

Substituting \(r = 2\), we get:

\(T_2 = \binom{6}{2} x^{6-2} \left( -\frac{2}{x} \right)^2 = 15 \cdot x^4 \cdot \frac{4}{x^2} = 60x^2\).

Now, consider the product:

\(2 \cdot (-160) + \frac{3}{x^2} \cdot 60x^2 = -320 + 180 = -140\).

Thus, the term independent of \(x\) is \(-140\).

(i) The binomial expansion of \((1 - 2x)^5\) is given by:

\((1 - 2x)^5 = \sum_{k=0}^{5} \binom{5}{k} (1)^{5-k} (-2x)^k\)

For \(x^4\), \(k = 4\):

Coefficient = \(\binom{5}{4} (1)^{1} (-2)^4 = 5 \times 16 = 80\)

For \(x^5\), \(k = 5\):

Coefficient = \(\binom{5}{5} (1)^{0} (-2)^5 = 1 \times (-32) = -32\)

(ii) In the expansion of \((1 + px)(1 - 2x)^5\), the term in \(x^5\) is:

\((-32) + 5p(1)(-2)^4 = 0\)

\(-32 + 80p = 0\)

\(80p = 32\)

\(p = \frac{32}{80} = \frac{2}{5}\)

First, expand \((a + x)^5\) using the binomial theorem:

\((a + x)^5 = a^5 + 5a^4x + 10a^3x^2 + \ldots\)

We are interested in the terms that will contribute to \(x^2\) in the product \(\left( 1 - \frac{2x}{a} \right)(a + x)^5\).

Consider the term \(-\frac{2x}{a} \cdot a^5\) which contributes \(-\frac{2a^5x}{a} = -2a^4x\).

Now consider the term \(1 \cdot 10a^3x^2 = 10a^3x^2\).

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

\(-\frac{2}{a} \times 5a^4 + 10a^3 = -10a^3 + 10a^3 = 0\).

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