(a) Rationalise the denominator:

\[ z=\frac{3+\lambda i}{\lambda+2i}\times\frac{\lambda-2i}{\lambda-2i}. \]

\[ z=\frac{(3+\lambda i)(\lambda-2i)}{\lambda^2+4}. \]

Expanding the numerator,

\[ (3+\lambda i)(\lambda-2i)=3\lambda-6i+\lambda^2 i+2\lambda. \]

\[ =5\lambda+(\lambda^2-6)i. \]

So

\[ z=\frac{5\lambda}{\lambda^2+4}+\frac{\lambda^2-6}{\lambda^2+4}i. \]

For \(\arg z=\frac14\pi\), the real and imaginary parts must be equal and positive.

Therefore

\[ 5\lambda=\lambda^2-6. \]

\[ \lambda^2-5\lambda-6=0. \]

\[ (\lambda-6)(\lambda+1)=0. \]

So

\[ \lambda=6\quad\text{or}\quad \lambda=-1. \]

Since the real part must be positive, \(\lambda=6\).

Answer: \(\lambda=6\).

(b) When \(\lambda=6\),

\[ z=\frac{3+6i}{6+2i}. \]

Use

\[ \left|\frac{z_1}{z_2}\right|=\frac{|z_1|}{|z_2|}. \]

So

\[ |z|=\frac{|3+6i|}{|6+2i|}. \]

\[ |z|=\frac{\sqrt{3^2+6^2}}{\sqrt{6^2+2^2}} =\frac{\sqrt{45}}{\sqrt{40}}. \]

\[ |z|=\sqrt{\frac{9}{8}}=\frac{3\sqrt2}{4}. \]

Answer: \(\frac{3\sqrt2}{4}\).