= Solution
The lowest <singlet state> is
$$
\boxed{\Psi_{\rm s}=\frac1{2\pi}|0,0\rangle,\qquad E_{\rm s}=0,\quad g_{\rm s}=1}.
$$
For the lowest <triplet states> take $\lambda=\pm1$ and $m_s=-1,0,1$:
$$
\boxed{\Psi_{{\rm t},\lambda,m_s}(\phi_1,\phi_2)=
\frac{e^{i\lambda\phi_2}-e^{i\lambda\phi_1}}{2\pi\sqrt2}|1,m_s\rangle,\qquad
E_{\rm t}=\epsilon_R,\quad g_{\rm t}=6}.
$$
<orthogonality> of $u_0$ and $u_\lambda$ verifies <wavefunction normalization> of the antisymmetric spatial factor. At this same <energy> there are also two excited <singlet states>, obtained by replacing its minus sign with a plus sign and the <spin> factor by $|0,0\rangle$. Together they reproduce the eight states counted in part a.
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