Triplet and singlet states: fermionic or bosonic? Suppose we have two spin-1/2 particles with no orbital angular momentum. We choose to work with the eigenbasis of total angular momentum $S^2$ and $S_z$, which gives us the triplet and the singlet states:
$$ \begin{align}
(s=1, \rm triplet, \rm symmetric)  
\begin{cases}
&|11\rangle=|\uparrow \uparrow\rangle \\ &|10\rangle=\frac{1}{\sqrt{2}}\left(|\uparrow \downarrow\rangle + |\downarrow \uparrow \rangle\right) \\ &|1-1\rangle=|\downarrow \downarrow\rangle 
\end{cases} \\ \\
(s=0, \rm singlet, \rm antisymmetric)  
\begin{cases}
&|00\rangle=\frac{1}{\sqrt{2}}\left(|\uparrow \downarrow\rangle - |\downarrow \uparrow \rangle\right) 
\end{cases}
\end{align} $$


*

*Both the triplet and the singlet states have integer-valued total spins. This suggests the composite system of two spin-1/2 particles behaves bosonically. Although the triplet state respects this by being totally symmetric, the singlet state is totally antisymmetric. Given there are no other parts to the wavefunction for us to antisymmetraize, we are stuck with a totally antisymmetric state describing $s=0$, which is for bosons. What am I missing here that gives rise to this contradiction? 

*Are the four listed states above always allowed? Or, does it depend on whether the two spin-1/2 particles are identical or distinguishable? 
I'm thinking that if they are distinguishable, then all four states are allowed, keeping in mind my confusion described in Question 1 (that is, I think the system should behave bosonically, but the singlet state is antisymmetric).
If the particles are identical, then I cannot tell them apart, and as far as I can tell, I have a composite system of two fermions, and I know the composite state must be antisymmetric. Therefore, only the singlet state would be allowed. 
 A: *

*There is no contradiction. A particle's spin is not its only attribute. A two-fermion state must be antisymmetric with respect to the exchange of all of their attributes, not just spin. If the state is symmetric with respect to the exchange of their spins, then it is antisymmetric with respect to the excahge of their other attributes (such as location or momentum, not shown in the OP), and conversely.

*The four states shown are allowed whether the particles are distinguishable or "identical" (same species). For the identical-particle case, consider the two electrons in orthohelium and parahelium (https://en.wikipedia.org/wiki/Helium_atom). For the non-identical-particle case, consider the various possible states of a hydrogen atom, taking into account the parallel or antiparallel configuration of the electron/nucleon spin (http://www.feynmanlectures.caltech.edu/III_12.html).
A: First on your point 1. There is no contradiction between a two-spin system having an antisymmetric state with respect to exchange of its component parts, yet being symmetric with respect to exchange of that pair with some other pair. 
The two-spin system is, overall, bosonic, as you say. In particular, the singlet state has the correct properties to be a $S=0$ state. For example, it does not change under rotations of the coordinate system. 
One can take any number of pairs of fermions in such a state, and push them all into the same spatial state. If these fermions are all mutually indistinguishable then the state will be symmetric with respect to exchange of one pair with any other pair, and antisymmetric with respect to exchange of one fermion with any other fermion. 
Now point 2. A pair of fermions has both spin and spatial properties. Their joint state can sometimes be factorised into a spatial part in a tensor product with a spin part. This does not always happen. Whether or not the state can be so factorized, it has to be antisymmetric with respect to exchange of those two fermions if the fermions are a pair of the same type of particle (e.g. two electrons). If the state can be factorized then its overall antisymmetry is achieved if:
either the spin state is the singlet and the spatial state is symmetric
or the spin state is the triplet and the spatial state is antisymmetric
Therefore if either kind of spatial state is available to the system, then so is either kind of spin state. If the two particles are in the same spatial state, then the overall spatial state cannot help but be symmetric, so in that case the spin state must be the singlet.
