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

\(T_k = \binom{7}{k} (2)^{7-k} (ax)^k\).

For the coefficient of \(x\), set \(k = 1\):

\(T_1 = \binom{7}{1} (2)^6 (ax)^1 = 7 \times 2^6 \times a \times x\).

The coefficient of \(x\) is \(7 \times 2^6 \times a\).

For the coefficient of \(x^2\), set \(k = 2\):

\(T_2 = \binom{7}{2} (2)^5 (ax)^2 = 21 \times 2^5 \times a^2\).

The coefficient of \(x^2\) is \(21 \times 2^5 \times a^2\).

Equating the coefficients:

\(7 \times 2^6 \times a = 21 \times 2^5 \times a^2\).

Divide both sides by \(2^5\):

\(7 \times 2 \times a = 21 \times a^2\).

Divide both sides by \(a\) (since \(a \neq 0\)):

\(14 = 21a\).

Solve for \(a\):

\(a = \frac{14}{21} = \frac{2}{3}\).

To find the coefficient of \(x^5\) in the expansion of \(\left(x^2 - \frac{a}{x}\right)^7\), we use the binomial theorem.

The general term in the expansion is given by:

\(T_k = \binom{7}{k} (x^2)^{7-k} \left(-\frac{a}{x}\right)^k\)

We need the term where the power of \(x\) is 5:

\((x^2)^{7-k} \left(-\frac{a}{x}\right)^k = x^{2(7-k)} \cdot (-a)^k \cdot x^{-k} = x^{14-2k-k} = x^{14-3k}\)

Set the exponent equal to 5:

\(14 - 3k = 5\)

Solving for \(k\):

\(14 - 5 = 3k\)

\(9 = 3k\)

\(k = 3\)

Substitute \(k = 3\) into the general term:

\(T_3 = \binom{7}{3} (x^2)^4 \left(-\frac{a}{x}\right)^3\)

\(= \binom{7}{3} x^8 \cdot (-a)^3 \cdot x^{-3}\)

\(= \binom{7}{3} (-a)^3 x^5\)

The coefficient of \(x^5\) is:

\(\binom{7}{3} (-a)^3 = -280\)

Calculate \(\binom{7}{3}\):

\(\binom{7}{3} = 35\)

Thus:

\(35(-a)^3 = -280\)

\((-a)^3 = -8\)

\(a^3 = 8\)

\(a = 2\)

(a) To expand \(\left( x + \frac{2}{x} \right)^5\), use the binomial theorem:

\(\sum_{k=0}^{5} \binom{5}{k} x^{5-k} \left( \frac{2}{x} \right)^k\).

Calculate each term:

\(\binom{5}{0} x^5 = x^5\)

\(\binom{5}{1} x^4 \cdot \frac{2}{x} = 10x^3\)

\(\binom{5}{2} x^3 \cdot \left( \frac{2}{x} \right)^2 = 40x\)

\(\binom{5}{3} x^2 \cdot \left( \frac{2}{x} \right)^3 = \frac{80}{x}\)

\(\binom{5}{4} x \cdot \left( \frac{2}{x} \right)^4 = \frac{80}{x^3}\)

\(\binom{5}{5} \left( \frac{2}{x} \right)^5 = \frac{32}{x^5}\)

Thus, the expansion is \(x^5 + 10x^3 + 40x + \frac{80}{x} + \frac{80}{x^3} + \frac{32}{x^5}\).

(b) Consider the expansion \((a + bx^2) \left( x + \frac{2}{x} \right)^5\).

The coefficient of \(x\) is zero:

\(40a + (\text{coefficient of } x^{-1})b = 0\)

The coefficient of \(\frac{1}{x}\) is 80:

\((\text{coefficient of } x^{-1})a + (\text{coefficient of } x^{-3})b = 80\)

From part (a), the coefficient of \(x^{-1}\) is 80 and \(x^{-3}\) is 80.

Thus, the equations are:

\(40a + 80b = 0\)

\(80a + 80b = 80\)

Solving these equations:

From \(40a + 80b = 0\), we get \(a = -2b\).

Substitute into \(80a + 80b = 80\):

\(80(-2b) + 80b = 80\)

\(-160b + 80b = 80\)

\(-80b = 80\)

\(b = -1\)

Substitute \(b = -1\) into \(a = -2b\):

\(a = 2\)

Thus, \(a = 2\) and \(b = -1\).

To find the coefficient of \(x^3\) in the expansion of \((a + x)^5\), we use the binomial theorem:

\(\text{Coeff of } x^3 \text{ in } (a + x)^5 = \binom{5}{3} a^{5-3} x^3 = 10a^2\)

For \((2 - x)^6\), the coefficient of \(x^3\) is:

\(\text{Coeff of } x^3 \text{ in } (2 - x)^6 = \binom{6}{3} (2)^{6-3} (-1)^3 x^3 = -160\)

Adding these coefficients gives the total coefficient of \(x^3\):

\(10a^2 - 160 = 90\)

Solving for \(a\):

\(10a^2 = 250\)

\(a^2 = 25\)

\(a = 5\)

To find the coefficient of \(x^2\) in the expansion of \(\left( k + \frac{1}{3}x \right)^5\), we use the binomial theorem:

\(\binom{5}{2} k^{5-2} \left( \frac{1}{3}x \right)^2\)

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

\(\binom{5}{2} k^3 \left( \frac{1}{3} \right)^2\)

\(= 10 \times k^3 \times \frac{1}{9}\)

We know this coefficient is 30, so:

\(10 \times k^3 \times \frac{1}{9} = 30\)

\(\frac{10}{9} k^3 = 30\)

\(k^3 = 30 \times \frac{9}{10}\)

\(k^3 = 27\)

\(k = 3\)