= Solution
Spin-one particles are <boson>[bosons], so their total two-particle state must be symmetric. Write $|m,n\rangle=|m\rangle_1|n\rangle_2$ for $m,n\in\{1,0,-1\}$. A basis of symmetric spin states is
$$
|1,1\rangle,\quad |0,0\rangle,\quad |-1,-1\rangle,\quad
\frac{|1,0\rangle+|0,1\rangle}{\sqrt2},\quad
\frac{|1,-1\rangle+|-1,1\rangle}{\sqrt2},\quad
\frac{|0,-1\rangle+|-1,0\rangle}{\sqrt2}.
$$
Denote these six states by $\chi_1^+,\ldots,\chi_6^+$. A basis of antisymmetric spin states is
$$
\chi_1^-=\frac{|1,0\rangle-|0,1\rangle}{\sqrt2},
\quad
\chi_2^-=\frac{|1,-1\rangle-|-1,1\rangle}{\sqrt2},
\quad
\chi_3^-=\frac{|0,-1\rangle-|-1,0\rangle}{\sqrt2}.
$$
This is the six-plus-three decomposition of <two identical spin-one bosons>.
In the ground orbital both particles occupy $n=0$, so the spatial wavefunction
$$
\Phi_{00}(x_1,x_2)=\psi_0(x_1)\psi_0(x_2)
$$
is symmetric. The complete list of ground states is therefore
$$
\boxed{
\Phi_{00}\chi_r^+,
\qquad r=1,\ldots,6,
\qquad E=2E_0.}
$$
At the first excited energy, one particle occupies $n=0$ and the other $n=1$. The normalized symmetric and antisymmetric spatial states are
$$
\Phi_{01}^{\pm}(x_1,x_2)
=\frac{
\psi_0(x_1)\psi_1(x_2)
\pm\psi_1(x_1)\psi_0(x_2)
}{\sqrt2}.
$$
Overall bosonic symmetry permits symmetric spatial times symmetric spin and antisymmetric spatial times antisymmetric spin. Hence all first excited states are
$$
\boxed{
\Phi_{01}^+\chi_r^+\quad(r=1,\ldots,6),
\qquad
\Phi_{01}^-\chi_s^-\quad(s=1,2,3),
\qquad
E=E_0+E_1.}
$$
The ground and first excited degeneracies are therefore six and nine.
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