$\DeclareMathOperator\Ann{Ann}\DeclareMathOperator\Tr{Tr}$My answer is somewhat complementary to [Nik Weaver's](https://mathoverflow.net/a/382826), and admitedly more focused on Question 2 since I have nothing more to add to the latter regarding Question 1.

When you deal with a *not necessarily commutative* C*-algebra $\mathfrak{A}$, the Riesz representation theorem is no longer the tool one uses to represent states. What is usually done is the so-called *Gel'fand–Naimark–Segal (GNS) construction* of a *-representation $\pi_\omega:\mathfrak{A}\rightarrow\mathfrak{B}(\mathscr{H}_\omega)$ of $\mathfrak{A}$ in a Hilbert space $\mathscr{H}_\omega$ starting from a state $\omega$ on $\mathfrak{A}$.

Let us recall the GNS construction for convenience. We assume for now that $\mathfrak{A}$ has a unit $\mathbb{1}$. Recall that a state on $\mathfrak{A}$ is a linear map $\omega:\mathfrak{A}\rightarrow\mathbb{C}$ such that $\omega(a^*a)\geq 0$ for all $a\in\mathfrak{A}$ and $\omega(\mathbb{1})=1$. This implies that $$\mathfrak{A}\times\mathfrak{A}\ni(a,b)\mapsto\omega(a^*b)$$ is a positive semidefinite Hermitian sesquilinear form on $\mathfrak{A}$, and therefore satisfies the Cauchy–Schwarz inequality $$\lvert\omega(a^*b)\rvert^2\leq\omega(a^*a)\omega(b^*b)\leq\|a\|^2\omega(b^*b)\ ,\quad a,b\in\mathfrak{A}\ $$
(the latter inequality comes from the fact that $b=\|a\|^2\mathbb{1}-a^*a$ is a positive element of $\mathfrak{A}$ and hence has the form $b=c^*c$ for some $c\in\mathfrak{A}$) This means that the so-called *annihilator* $\Ann \omega$ of $\omega$ $$\Ann \omega=\{a\in\mathfrak{A} \mathrel\vert \omega(a^*a)=0\}$$ is a left ideal (hence a vector subspace) of $\mathfrak{A}$, consisting of the "zero (semi)norm" elements of $\mathfrak{A}$ with respect to the (semi)norm on $\mathfrak{A}$ induced by this sesquilinear form. Hence, the latter induces a complex scalar product on $\mathfrak{A}/\Ann \omega$ - if $[a],[b]$ are the respective equivalence classes of $a,b\in\mathfrak{A}$ modulo $\Ann \omega$, we write $$\langle[a],[b]\rangle=\omega(a^*b)\ .$$ Moreover, since $\Ann \omega$ is a left ideal, $\mathfrak{A}$ has a natural left action on $\mathfrak{A}/\Ann \omega$ as $$\pi_\omega(a)[b]=[ab]\ .$$ This defines a *-representation $\pi_\omega$ on $\mathfrak{A}/\Ann \omega$ which satisfies $$\|\pi_\omega(a)[b]\|\leq\|a\|\,\|[b]\|\ ,\quad a,b\in\mathfrak{A}\ ,$$ thanks to the Cauchy–Schwarz inequality. This means that $\pi_\omega$ extends uniquely to a *-representation of $\mathfrak{A}$ in the Hilbert space $\mathscr{H}_\omega=\overline{\mathfrak{A}/\Ann \omega}$ by bounded linear operators therein. The state $\omega$ is then represented in $\mathscr{H}_\omega$ by the unit-norm element $\Omega_\omega=[\mathbb{1}]$, for $\omega(a)=\langle\Omega_\omega,\pi_\omega(a)\Omega_\omega\rangle$ for all $a\in\mathfrak{A}$.

As you can see, the GNS construction "kind of" identifies $\mathfrak{A}$ with (a dense subspace of) the Hilbert space $\mathscr{H}_\omega$ - that is, modulo $\Ann \omega$. If the state is *faithful*, i.e. $\Ann \omega=\{0\}$, then $\mathfrak{A}$ is indeed identified with (a dense subspace of) $\mathscr{H}_\omega$ - this is equivalent to $\pi_\omega$ being *injective*, i.e. faithful. This happens regardless of $\mathfrak{A}$ having Hilbert-Schmidt elements or not. However, even though any nonzero C*-algebra possesses a good deal of faithful states (this is a consequence of the Hahn-Banach theorem), not all states of $\mathfrak{A}$ are faithful.

Depending on which kind of C*-algebra $\mathfrak{A}$ and reference state $\omega$ you have, all other states on $\mathfrak{A}$ *may* or *may not* be represented as trace-class operators $\rho$ in $\mathscr{H}_\omega$ with unit trace, that is, for all linear maps $\eta:\mathfrak{A}\rightarrow\mathbb{C}$ such that $\eta(a^*a)\geq 0$ for all $a\in\mathfrak{A}$ and $\eta(\mathbb{1})=1$ there *may* or *may not* be a $\rho_\eta\in\mathfrak{B}(\mathscr{H}_\omega)$ with $\Tr(\rho_\eta)=1$ such that $\eta(a)=\Tr(\rho_\eta\pi_\omega(a))$ for all $a\in\mathfrak{A}$.

A situation where this is true regardless of which $\omega$ you choose is when $\mathfrak{A}$ is finite dimensional (i.e. a ${}^*$-algebra of matrices). This remains true for $\mathfrak{A}=\mathfrak{B}(\mathscr{H})$ with a separable Hilbert space $\mathscr{H}$ (as mentioned at the end of [Nik Weaver's answer](https://mathoverflow.net/a/382826)), which is the case of the algebra of observables for physical systems with finitely many degrees of freedom. On the other hand, for the kind of C${}^*$-algebras that appear as algebras of observables for physical systems with *infinitely* many degrees of freedom (e.g. thermodynamic limits of quantum statistical systems and quantum field theory), this is usually *false* — as examples, one may cite thermal equilibrium states at different temperatures (in the thermodynamic limit), distinct pure thermodynamic phases of a non-pure thermal equilibrium state, different superselection sectors in quantum field theory, etc.

In physical terms, infinitely many degrees of freedom usually is a manifestation of *locality*. More precisely, one usually considers the self-adjoint elements of $\mathfrak{A}$ as *local* observables measured within certain space(-time) regions (or limits of Cauchy sequences thereof). This means that states $\eta$ of the form $\eta(a)=\Tr(\rho_\eta\pi_\omega(a))$ may be seen as states "accessible" from $\omega$ through "physically allowed local operations", at least to an arbitrary degree of accuracy. Any other state is seen as "disjoint" from $\omega$. In infinitely extended space(-time) regions, there are usually plenty of mutually disjoint states on $\mathfrak{A}$.