(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 \(b = -1\).