Interpretation of the electromagnetic field strength tensor as a spin-1 field If I understand correctly the electromagnetic field strength tensor $ F_{\mu\nu}$ could be considered as a spin-1 field. In that case, what can one say about the total spin and the $z$-component of the spin for this field? Also, how is $F_{\mu\nu}$'s spin related to that of the photon field ($A_{\mu}$)?  
 A: I think the existing answer is a pedagogical nightmare, which obfuscates the underlying truth: The spin of the Electromagnetic Field Strength tensor $F_{\mu\nu}$ is determined by its representation under the Lorentz group.
General Math:
$2$-Tensors live in the the tensor product space of a space and itself (that is what it means to be a tensor of rank $2$). If vectors live in the fundamental representation $V$, then $2$-tensors live in $V\otimes V$. It is a relatively simple mathematical fact that this 'square' tensor product can always be decomposed into Symmetric and Anti-Symmetric parts: $$V\otimes V \cong [V \odot V] \oplus [V \wedge V]$$
It is important to note that if $\operatorname{dim}(V) = n$:
$$\operatorname{dim}(V\otimes V) = n^2$$ and $$\operatorname{dim}(\operatorname{Sym}^2(V))= \frac{n(n+1)}{2},\quad \operatorname{dim}(\operatorname{Alt}^2(V)) = \frac{n(n-1)}{2}$$
That is to say, the space of $2$-tensors decomposes into symmetric tensors and anti-symmetric tensors. It is also well known that all representations of semi-simple lie groups appear as components of the direct sum within representations of tensor products of the fundamental representation. That is to say, because the tensor product space decomposes this way, each of those pieces of the direct sum must be a representation of the relevant group.
All of this is to say that Symmetric $2$-tensors and Anti-symmetric $2$-tensors, must represent fundamentally different kinds of physical quantities, because they are not the same dimension, and so cannot live in the same representation of the symmetry group our physics respects.
Specific to the physics: Representations of the Lorentz group can be indexed by pairs of half integers. $(j_1,j_2)$. The fundamental/vector representation can be identified with the $(\frac{1}{2},\frac{1}{2})$ rep. This is a spin-1 rep, because the spin of a rep is $j_1+j_2$, and the dimension is $(2j_1+1)(2j_2+1)$. The tensor product space in this case decomposes into the following direct sum:
$$
(\frac{1}{2},\frac{1}{2})\otimes(\frac{1}{2},\frac{1}{2}) = [(1,1)\oplus(0,0)]\oplus[(1,0)\oplus(0,1)] = [\operatorname{Sym}^2(V)]\oplus[\operatorname{Alt}^2(V)]
$$
Where I have intentionally grouped the representations into the symmetric and alternating parts. The way one should read the latter equation is: "The $16$ dimensional space of $2$-tensors on Minkowski space can be split into a $9$ dimensional Spin-$2$ traceless symmetric piece, the $1$-dimensional trace (transforms as a scalar because trace is invariant), and a $6$ dimensional Spin-$1$ anti-symmetric piece".
The Faraday tensor $F_{\mu\nu}$ is an anti-symmetric $2$-tensor. This means it must be a spin-$1$ object, because that is what anti-symmetric $2$-tensors are in special relativity. This is why the E and B fields behave like Euclidean vectors under rotations, but mix seemingly strangely under boosts.
All of this analysis can be done in the arena of spinors as well, as done by Stephen Blake, because the Tensor Product space of the bi-spinor representation of the Lorentz group also contains a copy of the representation of anti-symmetric tensors which we previously called $[(1,0)\oplus(0,1)]$. But I do not believe this is particularly enlightening when compared with starting in the more comfortable vector representation.
As for how the spins are 'related' it is merely that the Faraday tensor is constructed to be anti-symmetric tensor, from the potential vector $A^\mu$. Perhaps there is more to be said here about the exterior algebra and the action of the exterior derivative, but that is out of my wheelhouse. Lastly, the '$z$' component of the Faraday tensor does not particularly make sense, one needs two indices to specify a component of a $2$-tensor.
A: The spin of the electromagnetic field tensor $F_{\mu\nu}$ is best understood by writing it as a spinor. A spin 1 field is a represented by a symmetric spinor $\xi^{AB}$ or by a dotted symmetric spinor $\eta_{\dot{A}\dot{B}}$. In order to get the field transforming correctly under parity, the electromagnetic field has to be a direct sum using the symmetric spinor and it's complex conjugate dotted spinor.
\begin{equation}
F_{\mu\nu} \sim \xi^{AB}\oplus [\xi^{*}]_{\dot{A}\dot{B}}
\end{equation}
The symmetric spinor $\xi^{AB}$ has three independent complex components $\xi^{11},\xi^{12}=\xi^{21},\xi^{22}$. Linear combinations of these components correspond to the three complex components of the electromagnetic field $B^{r}+iE^{r}$. The source free Maxwell equations are obtained by acting on the symmetric spinor with the Hermitian momentum operator $\hat{p}^{\dot{A}}_{\ B}$
\begin{equation}
\hat{p}^{\dot{A}}_{\ B}\xi^{BC}=0
\end{equation}
This equation is similar to the Dirac equation for a massive spin 1/2 field. The photon is massless, so it has helicity = $\pm 1$ instead of spin (essentially spin 1 along or against the direction of flight). The photon has two helicity degrees of freedom, but the spin 1 field $F_{\mu\nu} \sim \xi^{AB}\oplus [\xi^{*}]_{\dot{A}\dot{B}}$ has six real components. The Maxwell equations $\hat{p}^{\dot{A}}_{\ B}\xi^{BC}=0$ project the spinor onto a two-dimensional subspace.
This is as far as I know how to answer the question at present. The gauge field $A^{\mu}$ is a four vector so it ought to be a Hermitian spinor field of type $X^{\dot{A}}_{\ B}$ which is the tensor product of two spin 1/2 fields. It has four components, so the gauge fixing must come in to reduce four to the two helicity degrees of freedom. 
