All half integer particles are antisymmetric by the virtue of the spin-statistics theorem, which states that half integer particles must follow Fermi-Dirac statistics. As for your doubt concerning the spatial part of the wave function is symmetric, the net wave function comprising of the spin part and the spatial wave part must be antisymmetric, and this allows the choice of the position part being symmetric with the spin part being antisymmetric. For our case we take the spatial part to be symmetric and the spin part to be antisymmetric.
Exchange interaction
Exchange interaction arises in identical particles being subject to exchange symmetry, that is, either remaining unchanged (symmetric) or changing its sign (antisymmetric) when two particles are exchanged. For identical bosons, this results in attraction, while it results in repulsion for identical fermions (that is, when they have symmetric total wave-functions). But when the wavefunction is antisymmetric, the fermions attract each other.
Why this arises
If you take the Hamiltonian of, lets say a two-electron atom (which can be taken as a simple model for the covalent bond, although not exact, but from which basic features about exchange interaction can be understood from it), then in addition to the Coulumb interaction between nucleus and the electron, there will be a self interaction term also. Treating this self interaction term by Time independent perturbation theory, we can find out that there are two energy eigenvalues for the system, corresponding to the symmetric and the antisymmetric states.
Now in order to include effects due to spin, we take into account the Slater determinant, and find out the eigenvalues for the potential energy, which is now represented by a matrix. The state which is favored more (parallel spins or antiparallel spins) depends on the exchange constant, which is dependent on the difference between the energy eigenvalues of the symmetric and the antisymmetric ones.
Now if the exchange constant is positive, then the symmetric spin state is more preferred, while in the case of a negative exchange constant, the antisymmetric spin state is preferred. In our case of a two electron atom, the antisymmetric spin state is preferred, since it results in a lower energy than what would have been there if there was no spin/parallel spin. That indicates the reason to a good extent behind why covalent bonds are formed better if fermions are involved in the picture with antisymmetric spin wavefunctions, as can be inferred from the analysis of this two electron atom problem.
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.
Best Answer
Let's denote the spins of the individual particles by $j_1 = j_2 = I$, the quantum numbers for the $z$-components of their angular momentum by $m_1$ and $m_2$, the spin of their combined state by $J$, and the $z$-component of the angular momentum of the combined state by $M$. We have two bases for the states of these particles: the "individual particle" basis, denoted by $$ |I \, m_1 \, I \, m_2 \rangle $$ and the "combined particle" basis, denoted by $$ |J \, M \rangle. $$ (Note that $j_1$ and $j_2$ are also still "good" quantum numbers for this latter state; but including them in both notations is redundant and can confuse things, so I'll omit them.) Finally, let's denote by $\hat{E}$ the exchange operator between particles 1 & 2, i.e., we define $\hat{E}$ such that $$ \hat{E} |I \, m_1 \, I \, m_2 \rangle = |I \, m_2 \, I \, m_1 \rangle. $$
We can perform a basis transformation to express any state $|J \, M \rangle$ in terms of the basis $|I \, m_1 \, I \, m_2 \rangle$: $$ |J \, M \rangle = \sum_{j_1, m_1, j_2, m_2} |I \, m_1 \, I \, m_2 \rangle \langle I \,m_1 \,I \, m_2 | J \, M \rangle $$ The coefficients $\langle I \,m_1 \,I \, m_2 | J \, M \rangle$ are known as the Clebsch-Gordan coefficients. We want to know what happens when we apply the exchange operator $\hat{E}$ to this state: $$ \hat{E} |J \, M \rangle = \sum_{j_1, m_1, j_2, m_2} |I \, m_2 \, I \, m_1 \rangle \langle I \,m_1 \,I \, m_2 | J \, M \rangle = \sum_{j_1, m_1, j_2, m_2} |I \, m_1 \, I \, m_2 \rangle \langle I \,m_2 \,I \, m_1 | J \, M \rangle $$ (The second step is just a relabelling of dummy indices $m_1$ and $m_2$ in the sum.) We can see that $|J \, M \rangle$ will be an eigenstate of $\hat{E}$ if and only if all of the Clebsch-Gordan coefficients $\langle I \,m_1 \,I \, m_2 |J \, M \rangle$ are multiplied by the same factor when we exchange $m_1 \leftrightarrow m_2$. Thankfully, the Clebsch-Gordon coefficients satisfy the identity $$ \langle j_1 \,m_1 \, j_2 \, m_2 | J \, M \rangle = (-1)^{j_1 + j_2 - J} \langle j_2 \,m_2 \, j_1 \, m_1 | J \, M \rangle $$ and so we have \begin{multline} \hat{E} |J \, M \rangle = \sum_{j_1, m_1, j_2, m_2} |I \, m_1 \, I \, m_2 \rangle \left[ (-1)^{2I - J} \langle I \,m_1 \,I \, m_2 | J \, M \rangle \right] \\ = (-1)^{2I - J} \sum_{j_1, m_1, j_2, m_2} |I \, m_1 \, I \, m_2 \rangle \langle I \,m_1 \,I \, m_2 | J \, M \rangle = (-1)^{2I - J} |J \, M \rangle. \end{multline} Thus, the combined states are symmetric when $2I$ and $J$ are both even or both odd, and antisymmetric when one quantity is even and the other is odd.