0) Your guess about the Hilbert space is on the right track, but not correct. The space of gauge invariant operators is much too big; you have to mod out by the equations of motion in an appropriate sense. (Think about the case of 1d quantum mechanics, where the gauge symmetry is trivial. The Hilbert space is $L^2(\mathbb{R})$, generated by time zero position observables, not $L^2(\mathbb{R}^{\mathbb{R}})$, which is what you'd get if you used all observables.)
1) In pure Yang-Mills, the Wilson loops are a complete set of observables. (Credit, iirc, goes to Migdal.)
2) You can recover the local observables by taking the limit of small loops.
3) It's not always true that observables are gauge invariant polynomials in the fields. Wilson loops aren't polynomials!
4) If there's enough matter fields, the gauge theory may not confine, in which case, baryons and mesons are not the only observables. How many matter fields are required depends on the gauge group.
5) Not sure what you're asking here. Why do you think they should be baryons or mesons?
I disagree with Bruce Lee's answer. Matter fields can lie in any representation of the gauge group's Lie algebra - what representation they lie in needs to be specified when defining the theory. (Well, technically matter fields don't lie in the representation itself, but rather in the vector space that the representation acts on.) For example, in the Standard Model, both the Higgs field and the Weyl field $l$ containing the neutrino field and the $e$ part of the charged lepton lie in the fundamental rep of $\mathfrak{su}(2)$ but the trivial rep of $\mathfrak{su}(3)$, while the $\bar{e}$ part of the charged lepton lies in the trivial rep of both $\mathfrak{su}(2)$ and $\mathfrak{su}(3)$. Similarly, the $q$ field containing the $u$ and $d$ parts of the quarks lies in the $3$ (fundamental) rep of $\mathfrak{su}(3)$ and the fundamental rep of $\mathfrak{su}(2)$, while the $\bar{u}$ and $\bar{d}$ parts of the quarks lie in the $\bar{3}$ (complex conjugate of the fundamental) rep of $\mathfrak{su}(3)$ but the trivial rep of $\mathfrak{su}(2)$.
The gauge fields themselves lie in different representations of the gauge group's Lie algebra in different terms in the Lagrangian, depending on what they are coupled to. In the gauge fields' kinetic term $-\frac{1}{2} \text{Tr} \left( F^{\mu \nu} F_{\mu \nu} \right)$, they lie in the fundamental rep. In terms that take the form $D_\mu \varphi D^\mu \varphi$ where the gauge fields couple to matter fields via the covariant derivative, they lie in the same rep as the matter fields to which they are coupled.
Now assume the gauge field lies in the fundamental representation, and consider an infinitesimal gauge transformation $U = I - i g \theta(x) + o(\theta^2)$. The scalar field $\theta_{ij}(x)$ is a linear operator in the fundamental rep of the Lie algebra. But if we decompose it in a particular basis of generators $\{ T^a_{ij} \}$ as $\theta_{ij}(x) = \phi_a(x) T^a_{ij}$, then each component of the new field $\phi_a(x)$ specifies one generator's weight in the infinitesimal transformation. $\phi(x)$ therefore lies in the vector space acted on by the Lie algebra's adjoint representation, so mathematically it looks just like a matter field in the adjoint representation. The gauge field transforms as $A_\mu(x) \to A_\mu(x) - D_\mu \phi(x)$, where the gauge field outside the covariant derivative is in the fundamental rep, but the gauge field inside the covariant derivative lies in the adjoint rep (because it's coupled to a scalar in the adjoint rep).
Edit: upon further reflection, which representation the gauge field lies in depends on how you think about it. If you expand out the covariant derivative, then the term that couples the gauge and matter fields always looks like
$$g\, \bar{\Psi}_i(x)\, \gamma^\mu\, A_\mu^a(x)\, (T_R)^a_{ij}\, \Psi_j(x),$$ where I've suppressed the spinor indices. $T_R$ represents the generators of the gauge group in the representation $R$. The index $a$ always runs over the different generators - or equivalently, it runs over the indices of the adjoint-representation vector space. The indices $i$ and $j$ run over the basis vectors of the vector space corresponding to the representation $R$. There are two different valid ways of thinking about this interaction term: you can consider the gauge field itself to just be $A_\mu^a(x)$. Under this definition, the gauge field always lies in the vector space corresponding to the adjoint representation, and the tensor $(T_R)^a_{ij}$ is like a three-point coupling between the matter fields and the gauge field. I believe that these are the more traditional words to use to describe the interaction.
But there's another way to think about it, which is the approach that Srednicki's QFT textbook uses and the way I used above: the field $A_\mu^a(x)$ is always contracted with the generator matrices $(T_R)^a_{ij}$ in some representation. (Although sometimes this contraction isn't explicit, because the generator matrices are implicitly "hidden" in the invariant symbols $f^{abc}$ or $d^{abc}$, which are defined in terms of them.) So it's arguably more straightforward to think of the contracted quantity $$(A_{R\mu})_{ij}(x) := A_{\mu}^a(x)\, (T_R)^a_{ij}$$ as the fundamental field - one fewer index to worry about. (In this way of thinking, the fields $A_{\mu}^a(x)$) are simply the components of the fundamental field $(A_\mu^R)_{ij}(x)$ in a particular basis of generators.) Under this definition, the gauge field no longer lies in the vector space of the adjoint representation, but instead it's an operator on the vector space of the arbitrary representation $R$. The two ways of thinking are completely equivalent, they only differ in which side of the definition above is considered to be the gauge field itself and which side is a quantity derived from the gauge field. But in the second way of thinking, the gauge field operators act on different representations of the gauge Lie algebra in different coupling terms of the Lagrangian, and therefore in terms where the representation is not faithful, some of its degrees of freedom get projected out. Some people find this distasteful.
Things are a bit different for the kinetic term $\mathcal{L}_\text{kin} = -\frac{1}{4} F^{a \mu \nu} F_{a \mu \nu}$ (under the first convention, where the $A_\mu^a$ color-vectors are considered the primary fields) or $\mathcal{L}_\text{kin} = -\frac{1}{2} \text{Tr}(F_{\mu \nu} F^{\mu \nu})$ (under the second convention, where the $(A_\mu)_{ij}$ color-operators are considered the primary fields). It turns out that here you don't even need to specify a representation at all - you only need the information contained in the abstract Lie algebra structure, where the elements of the Lie algebra aren't thought of as linear operators with a commutator structure, but just as vectors with an abstract Lie bracket structure. The Lie bracket structure is enough to define a symmetric, bilinear inner product on the abstract Lie algebra called the (negative of the) Killing form. When we say "Trace" in the kinetic term, we really mean this Killing form - we don't need to actually take the trace of any linear operator in any particular representation. So the kinetic term is completely representation-independent. (We typically require the gauge group to be compact and semisimple in order for the Killing form to have definite signature, so that the kinetic term is bounded below.)
Best Answer
It's a lot of questions but they have pretty easy answers, so here they are:
Gauss's law is just $\mbox{div }\vec D=\rho$ in electrodynamics. Note that it contains no time derivatives so it's not really an equation describing evolution: it's an equation restricting the allowed initial conditions. More generally, it's the equation of motion that you get by varying the Lagrangian with respect to $A_0$, the time component of the gauge field, so the corresponding equation of motion counts the divergence of the electric field minus the electric sources (charge density) that must be equal to it. This difference is nothing than the generator of the overall $U(1)$ group, or any other group, if you consider more general theories, so the classical equation above is promoted to the quantum equation in which $(\mbox{div }\vec D-\rho)|\psi\rangle=0$ which just means that the state $|\psi\rangle$ is gauge-invariant.
The traces you are encountering here - in non-Abelian theories - are just traces over fundamental (or, less frequently, adjoint) indices of the Yang-Mills group. They're different than traces over the Hilbert space. You must distinguish different kinds of indices. Tracing over some color indices doesn't change the fact that you still have operators.
"Adjoint of a gauge group" is clearly the same thing as "Adjoint representation of the Lie group that is used for the gauge group."
It's not true that all compact spaces eliminate all massless fields - for example, the Wilson line of a gauge field remains a perfectly massless scalar field on toroidal compactifications in supersymmetric theories - but in most other, generic cases, it's true that compactification destroys the masslessness of all fields. All the Fourier (or non-zero normal) components of the fields that nontrivially depend on the extra dimensions - the Kaluza-Klein modes - become massive because of the extra momentum in the extra dimensions. But even the "zero modes" become massive in general theories because of the Casimir-like potentials resulting from the compactification.
Basic excitations are not "the same thing" as one-particle states. In fact, we want to use the word "excitation" exactly in the context when the goal is to describe arbitrary multi-particle states. But the basic excitations are just the creation operators (and the corresponding annihilation operators) constructed by Fourier-transforming the fields that appear in the Lagrangian, or that are elementary in any similar way.
You haven't constructed any "specific states above" so I can't tell you how some other states you haven't described are related. In this respect, your question remained vague. All of them are some states with particles on the Hilbert space - but pretty much all states may be classified in this way.
A mode of a quantum field is the term in some kind of Fourier decomposition, or - for more general compactifications and backgrounds - another term (such as the spherical harmonic) that is an eigenstate of energy i.e. that evolves as $\exp(E_n t/i\hbar)$ with time. So for example, a field $X(\sigma)$ for a periodic $\sigma$ may be written as the sum below. The individual terms for a fixed $n$ - or the factor $X_n$ or the function that multiplies it (this terminology depends on the context a bit) - are called the modes. $$\sum_{n\in Z} X_n \,\exp(in\sigma - i|n|\tau)$$
Gauge symmetry has to commute with the Hamiltonian because we want to ban the gauge-non-invariant states, and by banning them in the initial state, they have to be absent in the final state, too. So it has to be a symmetry. On the other hand, the representation theory is trivial because, as I said at the very beginning, we require physical states to be gauge-invariant; that was the comment about Gauss's law: states have to be annihilated by all operators of the type $\mbox{div }\vec D-\rho$ which are just generators of the gauge group at various points. In other words, all of the physical states have to be singlets under the gauge group. That's why the term "symmetry" is somewhat misleading: some people prefer to call it "gauge redundancy". So the Hilbert space is surely completely reducible - to an arbitrary number of singlets. You may reduce it to the smallest pieces that exist in linear algebra - one-dimensional spaces.
I just explained you again why all the physical states are singlets under the gauge group. The fact that $n$-particle states with identical particles are completely symmetric or completely antisymmetric reduces to the basic insight that in quantum field theory, particles are identical and their wave function has to be symmetric or antisymmetric (for bosons and fermions) because the corresponding creation operators commute (or anticommute) with each other.