To find the coefficient of \(x\) in the expansion of \(\left( \frac{1}{ax} + 2ax^2 \right)^5\), we use the binomial theorem. The general term in the expansion is given by:

\(T_k = \binom{5}{k} \left( \frac{1}{ax} \right)^{5-k} (2ax^2)^k\)

We need the term where the power of \(x\) is 1. The power of \(x\) in the general term is:

\(-5 + k + 2k = -5 + 3k\)

Setting \(-5 + 3k = 1\), we solve for \(k\):

\(3k = 6\)

\(k = 2\)

Substitute \(k = 2\) into the general term:

\(T_2 = \binom{5}{2} \left( \frac{1}{ax} \right)^{3} (2ax^2)^2\)

\(= 10 \times \frac{1}{a^3x^3} \times 4a^2x^4\)

\(= 10 \times \frac{4a^2x^4}{a^3x^3}\)

\(= 10 \times \frac{4a^2x}{a^3}\)

\(= \frac{40a^2x}{a^3}\)

\(= \frac{40x}{a}\)

The coefficient of \(x\) is \(\frac{40}{a}\). We are given that this coefficient is 5:

\(\frac{40}{a} = 5\)

Solving for \(a\):

\(40 = 5a\)

\(a = 8\)

To find the coefficient of \(x^3\) in the expansion of \((1 - 3x)^6 + (1 + ax)^5\), we consider the contributions from each binomial expansion separately.

For \((1 - 3x)^6\), the term containing \(x^3\) is given by the binomial coefficient:

\(\binom{6}{3} (1)^{6-3} (-3x)^3 = 20(-27x^3) = -540x^3.\)

For \((1 + ax)^5\), the term containing \(x^3\) is:

\(\binom{5}{3} (1)^{5-3} (ax)^3 = 10a^3x^3.\)

The total coefficient of \(x^3\) is:

\(-540 + 10a^3 = 100.\)

Solving for \(a\), we have:

\(10a^3 = 640\)

\(a^3 = 64\)

\(a = 4.\)

The term in \(x\) from \(\left(1 + \frac{x}{2}\right)^n\) is \(\frac{nx}{2}\).

In the expansion of \((3 - 2x)(1 + \frac{nx}{2} + \ldots)\), the coefficient of \(x\) is given by:

\(3 \cdot \frac{nx}{2} - 2 = 7\)

\(\frac{3n}{2} - 2 = 7\)

\(\frac{3n}{2} = 9\)

\(n = 6\)

Now, find the coefficient of \(x^2\):

The term in \(x^2\) from \(\left(1 + \frac{x}{2}\right)^n\) is \(\frac{n(n-1)}{2} \left(\frac{x}{2}\right)^2\).

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

\(\frac{3n(n-1)}{8} - \frac{2n}{2}\)

Substitute \(n = 6\):

\(\frac{3 \times 6 \times 5}{8} - 6 = \frac{90}{8} - 6 = \frac{45}{4} - 6 = \frac{21}{4}\)

The general term in the expansion of \((x + 2k)^7\) is given by:

\(T_r = \binom{7}{r} (x)^{7-r} (2k)^r\).

For the term in \(x^5\), set \(7-r = 5\), so \(r = 2\):

\(T_5 = \binom{7}{2} x^5 (2k)^2 = 21 \cdot 4k^2 x^5 = 84k^2 x^5\).

For the term in \(x^4\), set \(7-r = 4\), so \(r = 3\):

\(T_4 = \binom{7}{3} x^4 (2k)^3 = 35 \cdot 8k^3 x^4 = 280k^3 x^4\).

Since the coefficients of \(x^4\) and \(x^5\) are equal, we equate them:

\(84k^2 = 280k^3\).

Divide both sides by \(k^2\) (since \(k\) is non-zero):

\(84 = 280k\).

Solve for \(k\):

\(k = \frac{84}{280} = \frac{3}{10} = 0.3\).

To find the coefficient of \(x^2\) in the expansion of \((2 + ax)^6\), we use the binomial theorem:

\(\binom{6}{2} \cdot 2^4 \cdot (ax)^2 = 15 \cdot 16 \cdot a^2 x^2 = 240a^2 x^2\)

To find the coefficient of \(x^3\), we have:

\(\binom{6}{3} \cdot 2^3 \cdot (ax)^3 = 20 \cdot 8 \cdot a^3 x^3 = 160a^3 x^3\)

Setting the coefficients equal gives:

\(240a^2 = 160a^3\)

Solving for \(a\), we divide both sides by \(a^2\) (assuming \(a \neq 0\)):

\(240 = 160a\)

\(a = \frac{240}{160} = \frac{3}{2}\)