I'm using the standard notation throughout the whole answer.
The problem of adding angular momenta is essentially a change of basis, from one that diagonalizes $(S_1^2,S_2^2,S_{1z},S_{2z})$ to one that diagonalizes $(S^2,S_z,S_1^2,S_2^2).$ If you work out the problem which is given in many texts you will find the following transformation.
$$|s=1m=1,s_1=1/2 s_2=1/2\rangle =|++\rangle$$
$$|s=1m=0,s_1=1/2 s_2=1/2\rangle =2^{-1/2}[|+-\rangle+|-+\rangle]$$
$$|s=1m=-1,s_1=1/2 s_2=1/2\rangle =|--\rangle$$
$$|s=0m=0,s_1=1/2 s_2=1/2\rangle =2^{-1/2}[|+-\rangle-|-+\rangle]$$
The allowed values for total spin are $s=1$ and $0$,while the allowed values of $s_z$ are $\hbar,0$ and $-\hbar$.
For a system of two spins 1/2 particles the wavefunction have the following possible forms
$$\Psi =
\left\{
\begin{array}{l}
\psi_a\chi_s \\
\psi_s\chi_a
\end{array}
\right.$$
subscript $s$ and is to denote symmetric and anti-symmetric.
The particle of mass $m$ in the box of length $L$ in 1D is solved by wavefunctions
$$
\begin{align}
\psi_{n\alpha}&=A\sin (k_n x) e^{-\omega_n t}|\alpha \rangle\;, \\
k_n&=\frac{n\pi}{L}\;,\\
E_n&=\hbar \omega_n\;,\\
\omega_n&=\frac{\pi h n^2}{4L^2m}\;.
\end{align}
$$
Here, $|\alpha \rangle $ represents the spin state.
$$\Psi_{n\alpha m\beta}(x_1,x_2,t)=\psi_{n\alpha}(x_1,t)\psi_{m\beta}(x_2,t) - \psi_{m\beta}(x_1,t)\psi_{n\alpha}(x_2,t)$$
The energy of state $\Psi_{n\alpha m\beta}(x_1,x_2,t)$ can be calculated as
$$(H_1+H_2)\Psi_{n\alpha m\beta}(x_1,x_2,t)=(E_n+E_m)\Psi_{n\alpha m\beta}(x_1,x_2,t)$$
since each of the one-particle Hamiltonians acts on the respective one-particle wavefunction $\psi_{n\alpha}(x_1,t)$, which yields its eigenenergy $E_n$.
First consider state for which $\alpha=\uparrow$ and $\beta=\uparrow$.
The ground state is the lowest-lying energy state of the system. In this case, it would correspond to $\Psi_{1\uparrow 1\uparrow}$, but this function is identically zero. Then next two lowest-lying states are $\Psi_{1\uparrow 2\uparrow}$ and $\Psi_{2\uparrow 1\uparrow}$.Thanks to the antisymmetrization, $\Psi_{1\uparrow 2\uparrow} = -\Psi_{2\uparrow 1\uparrow}$ and it represents the ground state of the system with energy $E_1+E_2$. So the first two lowest energies are $$E^0=E^1=5E_0$$
For opposite spins, we choose $\alpha=\uparrow$ and $\beta=\downarrow$. Here, the lowest lying energy state is $\Psi_{1\uparrow 1\downarrow}$ and it has energy $2E_0$.
You may wonder because this doesn't match with the answer in the textbook, So the only thing I can conclude that there is a mistake in the problem or in solution. I hope this will help you. Best wishes!
Best Answer
What I get from the question is you want to find the wave functions of the ground and first excited states of a system of two non-interacting spin-1 bosons in an infinite square well potential.
Because the particles are non-interacting, you get the usual solutions for the infinite square well with quantized eigenvalues $E_n^{\sigma} = \frac{\hbar^2\pi^2}{2ma^2}n^2, \ n \in \{1,2,...\}, \ \sigma \in\{ \uparrow, 0, \downarrow\}$, degenerate in the spin index so $\psi_{n_i}(\vec{x}, \sigma) = \phi_{n_i}(\vec{x})\chi(\sigma)$ and the total wave function can be written as $$ \Psi_{n_1,n_2}(\vec{x}_1, \sigma_1; \vec{x}_2, \sigma_2) \equiv \Phi_{n_1,n_2}(\vec{x}_1,\vec{x}_2) X(\sigma_1, \sigma_2). $$ Since we have bosons, $\Psi$ must be symmetric under permutation of $(\vec{x}_1,\sigma_1)$ and $(\vec{x}_2, \sigma_2)$, so each $\Phi$ and $X$ must be either symmetric or anti-symmetric, as long as their product is symmetric.
For the ground state, the lower energy is clearly achieved with $n_1 = n_2 = 1$ and $\Phi_{1,1}(\vec{x}_1,\vec{x}_2) = \prod_{i\in\{1,2\}} \phi_1(\vec{x}_i)$ is symmetric, so $X$ must be as well. For the first excited state, you need $n_1 (n_2) = 1, \ n_2 (n_1) = 2$. You can construct $\Phi$ to be either symmetric or anti-symmetric, like $$ \Phi^{\pm}_{1,2}(\vec{x}_1,\vec{x}_2) = \frac{1}{\sqrt{2}}\left(\phi_1(\vec{x}_1)\phi_2(\vec{x}_2) \pm \phi_1(\vec{x}_2)\phi_2(\vec{x}_1) \right), $$ so the corresponding spin part has to be symmetric or anti-symmetric.
Now, the spin states can be expressed in any basis because the energy is independent of the spins. The basis of eigenvectors of the total spin are already in described in either symmetric or anti-symmetric, so it's a natural basis to take.
Taking a Clebsch-Gordan table, you can see that, as you said, total spin $s \in \{0,2\}$ are symmetric while $s = 1$ is anti-symmetric. The reason for this is that when combining two identical particles of spin $j$, the highest spin you can create is both particles maximally aligned $X(s=2j, m_s=2j) = X(m_1=j, m_2 = j)$, which is symmetric. To create the other vector of the subset $X(s = 2j, m_s)$, you apply $J_-$, a symmetric operator. on both sides, so all the states generated are symmetric.
To construct the next lower subset (here $X(s=2j-1, m_s)$), you take $$ X(s=2j, m_s=2j-1) = \frac{1}{\sqrt{2}}\left( X(m_1=j, m_2=j-1) + X(m_1=j-1, m_2=j) \right) $$ and $X(s=2j-1, m_s=2j-1)$ must be made of the same two vectors, but be orthogonal, so the only way is to take the anti-symmetric version with a $-$ instead of a $+$, so the subset $X(s=2j-1, m_s)$ is anti-symmetric and so on.
So for your ground state, you need $X$ symmetric so any $X(s \in \{0,2\}, m_s)$ and for your first excited state, either $X$ symmetric if $\Phi$ is symmetric, or any anti-symmetric $X(s=1,m_s)$ if $\Phi$ is anti-symmetric.