To answer your first question:
Particles and fields are separate. Particles are the irreducible excitations of fields. You can only get particles after quantizing fields.
However, you might often see people using particles without quantization of fields (in classical mechanics and GTR). You must understand that these are approximate models obtained by assuming the energy density of fields is concentrated in point like particles.
At the heart of non-quantized physics, we have continuous material fields for photons, electrons, quarks etc. These fields are (most generally tensor fields) of the form $${{\psi}_{(i)}}^{s...u}_{e....g}(x,y,z,t)$$ $(i)$ denotes the type of field (like photon, higgs boson etc.)These include scalars, vectors, co-vectors, spinors, etc. The Lagrangian density $L$ is usually function of the components of these various fields and the metric tensor. One needs to rely on observation and other considerations (like gauge symmetries) to construct a covariant (value always fixed at an event) Lagrangian. So the 'configuration' you speak of depends on all these factors.
Your second question is actually far more interesting. There are 2 SE tensors commonly used used.They differ by the divergence of an antisymmetric tensor. This paper: http://authors.library.caltech.edu/19366/1/GoMa1992.pdf discusses this in detail.
The first is the canonical SE tensor derived as a conserved current using Noether's theorem from the space-time translational invariance of the Lagrangian.
The second type of tensor is derived from considerations of diffeomorphic invariance of the action. It is called Belinfante - Rosenfeld SE tensor. A diffeomorphism is a very sophisticated and generalized notion of a translation. Let vector field $X$ be a generator of general diffeomorphism ${\phi}^*$ and volume of integration is $F$. $X$ vanishes outside $F$. So we have
$${\int}_{F}L{\eta}-{\phi}^*(L{\eta}) = 0$$
Thus
$${\int}_{F}D_{X}(L{\eta})=0$$ where ${\eta}$ is the volume form (I have suppressed the factor of 1/4!)
Expanding the RHS, we get:
$${\int}_{F}D_{X}(L{\eta}) = {\int}_{F}[(\frac{{\partial}L}{{\partial}{{{\psi}_{(i)}}^{s...u}_{e....g}}}-(\frac{{\partial}L}{{\partial}{{{\psi}_{(i)}}^{s...u}_{e....g;c}}})_{;c})D_{X}{{{\psi}_{(i)}}^{s...u}_{e....g}} + \frac{1}{2}T^{ab}D_{X}g_{ab}]{\eta}=0 $$
As you can see the first term is the Euler-Equation which is equal to zero for every field component, so each term in the first part of the integral vanishes.
Now a basic result that can be directly inferred from the definition of a diffeomorphism is $$D_{X}g_{ab}= X_{a;b} + X_{b;a}$$
Substituting this in the above formula
$${\int}_{F}D_{X}(L{\eta}) = {\int}_{F}((T^{ab}X_a)_{;b}-(T^{ab}_{;b})X_a){\eta}=0 $$
The first term can be transformed into a surface integral on the boundary of $F$ and vanishes as $X$ vanishes on the boundary of $F$. This leaves us with $${\int}_{F}D_{X}(L{\eta}) = {\int}_{F}-(T^{ab}_{;b})X_a{\eta}=0 $$
Now the above must always be true for any arbitrary $X$, this is only possible if $(T^{ab}_{;b})=0$ .
This tensor is always symmetric and gauge invariant, so it is far more useful in GTR than the canonical tensor. Refer to the paper linked above to know more details about the subtle differences between the two.
References:
Chapter 3, 'Large Scale Structure of Space-Time' by Hawking and Ellis
You should think at the way the Noether current is obtained. When an infinitesimal symmetry transformation is made spacetime dependent, that is the parameters $\omega^a$ that control the symmetry are taken as functions of the spacetime point $\omega^a=\omega^a(x)$, the action is not longer left invariant
$$
\delta S=-\int d^D x\, J^{a\,\mu}(x)\partial_\mu \omega^a(x)
$$
but rather it provides the definition of the current $J^{a\,\mu}$ that is conserved on-shell.
Now, let's look at the case of the energy momentum tensor: in this case, the translations $x^\nu\rightarrow x^\nu+\omega^\nu$ are made local $x^\nu\rightarrow x^\nu+\omega^\nu(x)$ so that
$$
\delta S=-\int d^D x\, T_\nu^\mu(x)\,\partial_\mu \omega^\nu(x)\,.
$$
Actually, one looks for a symmetric tensor $T^{\mu\nu}=T^{\nu\mu}$ so that one can rewrite the expression above in the following form
$$
\delta S=-\frac{1}{2}\int d^D x\, T^{\nu\mu}(x)\,(\partial_\mu \omega_\nu(x)+\partial_\nu\omega_\mu)\,.
$$
Now, here is the catch: if we were to transform the spacetime metric $g_{\mu\nu}$ (equal to $\eta_{\mu\nu}$ in the case at hand) as if $x^\nu\rightarrow x^\nu+\omega^\nu(x)$ was just an infinitesimal change of coordinates, that is
$$g_{\mu\nu}\rightarrow g_{\mu\nu}-(\partial_\mu \omega_\nu(x)+\partial_\nu\omega_\mu)\,,$$
then the action (rendered coordinates independent by the inclusion of the metric in the usual way such as $d^D x\rightarrow d^D x \sqrt{|g|}\,,\ldots$) would be left invariant
$$
\delta S=-\frac{1}{2}\int d^D x\, (\partial_\mu \omega_\nu(x)+\partial_\nu\omega_\mu)\left. \left(\sqrt{|g|}\,T^{\mu\nu}(x)+2\frac{\delta S(x)}{\delta g_{\mu\nu}}\right)\right|_{g_{\mu\nu}=\eta_{\mu\nu}}=0\,.
$$
From this equation it follows that the current associated with spacetime translations can be written as
$$
T^{\mu\nu}=\left. -\frac{2}{\sqrt{|g|}}\frac{\delta S}{\delta g_{\mu\nu}}\right|_{g_{\mu\nu}=\eta_{\mu\nu}} \qquad \mbox{evaluated on the bkg }g_{\mu\nu}=\eta_{\mu\nu}\,.
$$
It should be apparent that this definition gives a symmetric energy-momentum tensor that matches the one appearing the Einstein's equations. From the derivation above it should also be clear that the alternate versions of $T_{\mu\nu}$ arise because the definition of $T^{\mu\nu}$ via the variation of the action when the translation is made spacetime dependent does not uniquely fix it. For example, given a valid $T_{\mu\nu}$, one can always define another one $T^{\mu\nu}\rightarrow T^{\mu\nu}_B=T^{\mu\nu}+\partial_\rho B^{\rho\mu\nu} $ with an arbitrary $B^{\rho\mu\nu}=-B^{\mu\rho\sigma}$ which also gives
$$
\delta S=-\int d^D x\, T^{\mu\nu}(x)\partial_\mu\omega_\nu(x)=-\int d^D x\, T_B^{\mu\nu}(x)\partial_\mu\omega_\nu(x)
$$
up to an integration by parts. Einstein's equations break this degeneracy and nicely identify ``the'' energy-momentum tensor.
Best Answer
Well, firstly, we have to assume Lorentz covariance and general covariance of the theory. For non-relativistic theories all bets are off. Secondly, in case of fermions, one needs to generalize the Hilbert SEM tensor from variation wrt. a metric to a variation wrt. a vielbein, see e.g. my Phys.SE answer here. Then the generalized Hilbert SEM tensor is the canonical SEM plus the Belinfante-Rosenfeld improvement term. A proof is sketched in my Phys.SE answer here and links therein.
No, a symmetric SEM tensor is not unique. It is in principle possible to add improvement terms that respects the symmetry and the conservations laws.