The answer to the first question is **yes**. This follows from the following more general result.

**Terminology I: Ordered Banach spaces.** By a *pre-ordered Banach space* I mean a pair $(X,X_+)$ where $X$ is a real Banach space and $X_+$ is a non-empty closed subset of $X$ such that $X_+ + X_+ \subseteq X_+$ and $\alpha X_+ \subseteq X_+$ for each scalar $\alpha \ge 0$ (in other words: $X_+$ is a so-called *wedge* in $X$.)

The *dual wedge* of $X_+$ is the wedge
$$
X'_+ := \{x' \in X': \, \langle x',x \rangle \ge 0 \text{ for each } x \in X_+\}.
$$
Note that $(X', X'_+)$ is a pre-ordered Banach space, too. Moreover, for each $x \in X$ it follows from the Hahn-Banach theorem that $x \in X_+$ if and only if $\langle x', x\rangle \ge 0$ for each $x' \in X'_+$.

By iterating this procedure, one can also define the bi-dual wedge $X''_+$ of $X_+$ in $X''$.

**Terminology II: Polars** Let $\langle X,Y\rangle$ be a dual pair of two real vector spaces; in other words, $\langle \cdot, \cdot \rangle: X \times Y \to \mathbb{R}$ is a bi-linear map such that $X$ separates $Y$ and $Y$ separates $X$ via this map.

For every subset $A \subseteq X$ the subset
$$
A^\circ := \{y \in Y: \, \langle x, y \rangle \le 1 \text{ for all } x \in A \}
$$
of $Y$ is called the *polar* of $A$ in $Y$. Similarly, for each set $B \subseteq Y$ the subset
$$
{}^\circ B := \{x \in X: \, \langle x, y\rangle \le 1 \text{ for all } y \in B \}
$$
of $X$ is called the *polar* of $B$ in $X$.

Now, the **bi-polar theorem** (see for instance the theorem on page 126 in H. H. Schaefer's book "Topological vector spaces" (1971)) says the following:

**Theorem.** The so-called *bi-polar* $\left({}^\circ B \right)^\circ$ of a subset $B \subseteq Y$ is the closure of the convex hull of $B \cup \{0\}$ with respect to the topology on $Y$ induced by $X$ via the duality mapping $\langle \cdot, \cdot \rangle$.

Now we can apply this result to pre-ordered Banach spaces:

**Density of wedges in their bi-dual wedges** Let $(X,X_+)$ be a pre-ordered Banach space, and identify $X_+$ with a subset of $X''_+$ by means of evaluation.

**Theorem.** The wedge $X_+$ is weak${}^*$-dense in the bi-dual wedge $X''_+$.

*Proof.* We consider the dual pair $\langle X', X'' \rangle$ with respect to the usual duality. Then it is easily checked that the polar of $X_+ \subseteq X''$ in $X'$ equals the negative dual wedge $-X'_+$. Similarly, it is easy to see that the polar of $-X'_+$ in $X''$ equals the bi-dual wedge $X''_+$. Hence, the bi-polar theorem implies that $X''_+$ is the weak${}^*$-closure of $X_+$ in $X''$.

**Remark.** I believe that the same still works if we intersect the wedge with the unit ball, i.e. the intersection of $X_+$ with the unit ball is weak${}^*$-dense in the intersection of $X''_+$ with the unit ball. I have not checked the details, though.

**Application to the first question of the OP.** The space $B(\mathcal{H})$ is the complexification of the space of self-adjoint operators on $\mathcal{H}$. So to apply the general result above, one can choose $X$ to be the set of all those trace class operators that yield real values when applied to self-adjoint operators; then $X'$ is simply the self-adjoint part of $B(\mathcal{H})$, and $X''$ is the set of all bounded linear functionals on $B(\mathcal{H})$ that map self-adjoint operators to real values. The wedges $X_+$, $X'_+$ and $X''_+$ are the standard cones in these spaces. Since we have seen above that $X_+$ is weak${}^*$-dense in $X''_+$, this yields that desired result.