= Solution
A <separated dual pair> consists of real <vector spaces> $E,F$ and a <bilinear form> $\langle x,y\rangle$ that separates both variables: each nonzero $x\in E$ pairs nontrivially with some $y\in F$, and conversely. Thus $F$ embeds in the algebraic dual of $E$, and $E$ embeds in the algebraic dual of $F$.
The <weak topology> $\sigma(E,F)$ is the coarsest topology making all maps $x\mapsto\langle x,y\rangle$, $y\in F$, continuous. It makes $E$ a Hausdorff <locally convex space>, with topology generated by the <seminorms> $p_y(x)=|\langle x,y\rangle|$. A neighbourhood base at zero consists of finite intersections
$$
\{x:|\langle x,y_j\rangle|<\varepsilon_j,\ 1\leq j\leq m\},
\qquad\varepsilon_j>0.
$$
Separation ensures that the common zero set of all these seminorms is just zero, giving Hausdorffness.
We determine the <continuous dual of a weak topology>. Suppose a <linear functional> $\ell$ is continuous. A basic neighbourhood as above is contained in $\{x:|\ell(x)|<1\}$. If $x$ is in the common kernel of the finitely many evaluations, every scalar multiple of $x$ lies in that neighbourhood, so $\ell(x)=0$. Hence $\ell$ factors through the image of
$$
T:E\longrightarrow\mathbb R^m,\qquad Tx=(\langle x,y_1\rangle,\ldots,\langle x,y_m\rangle).
$$
Extend the resulting <linear functional> on $T(E)$ to $\mathbb R^m$. It has the form $z\mapsto\sum_j a_jz_j$, giving $\ell(x)=\langle x,\sum_ja_jy_j\rangle$. Conversely every such evaluation is continuous by the definition of the <weak topology>. Since the pairing separates $F$, this identification is injective, and
$$
\boxed{(E,\sigma(E,F))'=F.}
$$
For the remaining parts every topological assertion about $E'$ or its subsets uses the stipulated <weak-star topology> $\sigma(E',E)$.
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