= Solution
At small $k$,
$$
\frac d{dk}\log S_0(k)\bigg|_{k=0}
=-2i\left(\frac1{3\lambda}+\frac1{2\lambda}\right)
=-\frac{5i}{3\lambda}.
$$
Since $S_0=e^{2i\delta_0}$,
$$
\delta_0(k)=-\frac{5k}{6\lambda}+O(k^3).
$$
The scattering length and low-energy total cross-section are therefore
$$
\boxed{a_s=-\lim_{k\to0}\frac{\tan\delta_0(k)}k
=\frac5{6\lambda},
\qquad
\sigma_T\sim4\pi a_s^2=\frac{25\pi}{9\lambda^2}.}
$$
The poles at $k=2i\lambda$ and $k=3i\lambda$ lie on the positive imaginary axis. They represent two <bound states>, with energies
$$
E=-\frac{\hbar^2(2\lambda)^2}{2m},
\qquad
E=-\frac{\hbar^2(3\lambda)^2}{2m}.
$$
There are no resonance poles away from the imaginary axis in the lower half-plane. Positivity of $\lambda$ puts the poles on the physical bound-state axis, where $e^{ikr}$ decays; changing its sign would move them to the unphysical half-plane. This illustrates the <bound-state poles of a partial-wave S-matrix>.
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