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!