(a) Multiply both sides by \(a + 2i\) to clear the fraction:

\(2 + 3ai = \lambda(2 - i)(a + 2i)\).

Expand the right-hand side:

\(2 + 3ai = \lambda(2a + 4i - ai - 2)\).

Separate real and imaginary parts:

Real: \(2 = \lambda(2a - 2)\)

Imaginary: \(3a = \lambda(4 - a)\).

From the real part: \(2 = \lambda(2a - 2) \Rightarrow \lambda = \frac{2}{2a - 2}\).

From the imaginary part: \(3a = \lambda(4 - a) \Rightarrow \lambda = \frac{3a}{4 - a}\).

Equating the two expressions for \(\lambda\):

\(\frac{2}{2a - 2} = \frac{3a}{4 - a}\).

Cross-multiply and simplify:

\(2(4 - a) = 3a(2a - 2)\).

\(8 - 2a = 6a^2 - 6a\).

Rearrange to form a quadratic equation:

\(6a^2 - 4a - 8 = 0\).

Divide by 2:

\(3a^2 + 4a - 4 = 0\).

(b) Solve the quadratic equation \(3a^2 + 4a - 4 = 0\) using the quadratic formula:

\(a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 3, b = 4, c = -4\).

\(a = \frac{-4 \pm \sqrt{16 + 48}}{6}\).

\(a = \frac{-4 \pm \sqrt{64}}{6}\).

\(a = \frac{-4 \pm 8}{6}\).

\(a = \frac{4}{6} = \frac{2}{3}\) or \(a = \frac{-12}{6} = -2\).

For \(a = -2\), substitute back to find \(\lambda\):

\(3(-2) = \lambda(4 + 2) \Rightarrow -6 = 6\lambda \Rightarrow \lambda = -1\).

For \(a = \frac{2}{3}\), substitute back to find \(\lambda\):

\(3\left(\frac{2}{3}\right) = \lambda\left(4 - \frac{2}{3}\right) \Rightarrow 2 = \lambda\left(\frac{10}{3}\right) \Rightarrow \lambda = \frac{3}{5}\).