(a) Multiply numerator and denominator by the conjugate \(\lambda-2i\):

\[z=\frac{(3+\lambda i)(\lambda-2i)}{\lambda^2+4}=\frac{5\lambda+(\lambda^2-6)i}{\lambda^2+4}.\]

For argument \(\pi/4\), the real and imaginary parts must be equal and positive. Equating them gives

\[\lambda^2-6=5\lambda\quad\Longrightarrow\quad(\lambda-6)(\lambda+1)=0.\]

At \(\lambda=-1\), both parts are negative, giving argument \(-3\pi/4\). Reject this value. At \(\lambda=6\), both are positive, so \(\boxed{\lambda=6}\).

(b) Then \(z=(30+30i)/40=\frac34+\frac34i\), giving

\[\lvert z\rvert=\sqrt{\left(\frac34\right)^2+\left(\frac34\right)^2}=\boxed{\frac{3\sqrt2}{4}}.\]