= Solution
The <Hahn-Banach separation theorem for two convex sets> states that if $A$ and $B$ are disjoint nonempty convex subsets of a real <locally convex space> $X$ and $A$ is open, then there are a continuous linear functional $f\in X^*$ and a real number $c$ such that
$$
f(a)<c\leq f(b)
\qquad(a\in A, b\in B).
$$
To prove it, form
$$
C=A-B=\{a-b:a\in A, b\in B\}.
$$
This is open and convex and does not contain zero. The <separation of a point and an open convex set>, proved from the <Minkowski functional> and the <Hahn-Banach theorem>, gives a nonzero continuous $f$ with $f(c)<0$ for every $c\in C$. Hence $f(a)<f(b)$ for all $a,b$. Taking $c$ between $\sup_Af$ and $\inf_Bf$ gives the stated form.
For a <dual pair> $(E,F)$, the topology $\sigma(E,F)$ has a neighbourhood base at zero consisting of
$$
\{x:|f_j(x)|<\varepsilon, 1\leq j\leq n\},
\qquad f_j\in F.
$$
Every $f\in F$ is continuous by definition. Conversely, if a linear functional $g$ is continuous, some such neighbourhood lies in $\{|g|<1\}$. Therefore $\bigcap_j\ker f_j\subseteq\ker g$. Elementary linear algebra then gives $g\in\operatorname{span}\{f_1,\ldots,f_n\}\subseteq F$. Thus
$$
\boxed{(E,\sigma(E,F))^*=F}.
$$
For a normed space $X$, the weak topology is $\sigma(X,X^*)$; on $X^*$ the weak-star topology is $\sigma(X^*,X)$, using the canonical image of $X$ in $X^{**}$. If $X$ is reflexive, $X^{**}=J_X(X)$, so the weak and weak-star topologies on $X^*$ coincide. Conversely, if they coincide, every $x^{**}\in X^{**}$ is weak-star continuous. The dual-pair result says that every such functional is evaluation at some $x\in X$, so $J_X$ is onto and $X$ is reflexive.
For $A\subseteq E$, every $a\in A$ and zero satisfy every inequality defining $A^{\circ\circ}$, so $A\cup\{0\}\subseteq A^{\circ\circ}$. The latter is an intersection of weakly closed convex half-spaces, hence contains
$$
C=\overline{\operatorname{conv}}^{\sigma(E,F)}(A\cup\{0\}).
$$
If $x\notin C$, choose an open convex neighbourhood $V$ of zero with $(x+V)\cap C=\varnothing$. Applying the separation theorem to $x+V$ and $C$ gives $f\in F$ with
$$
\sup_{c\in C}f(c)<f(x).
$$
Because $0\in C$, the supremum is nonnegative. After multiplying $f$ by a positive scalar, $f\leq1$ on $C$ while $f(x)>1$. Thus $f\in A^\circ$ but $x\notin A^{\circ\circ}$. We conclude with the <Bipolar theorem for a dual pair>:
$$
\boxed{A^{\circ\circ}=\overline{\operatorname{conv}}^{\sigma(E,F)}(A\cup\{0\}).}
$$
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