Let $X$ be a Banach space over the complex field. Let $X^*$ denote its topological dual. If $S$ is a subspace of $X$, write $\mathrm{ann}_L(S)= \{ \varphi \in X^* : \varphi S = 0\}$ and note the result is a weak-star closed subspace of $X^*$. Similarly, if $S$ is a subspace of $X^*$, write $\mathrm{ann}_R (S) = \{ x \in X : S x = 0 \}$ and note the result is a weakly closed subspace of $X$. It is obvious that $\mathrm{ann}_L$ and $\mathrm{ann}_R$ are order-reversing. Also, if $S, T$ are subspaces of $X, X^*$ respectively, we have $ S \subset \mathrm{ann}_R(T) \Leftrightarrow T \subset \mathrm{ann}_L(S) \Leftrightarrow TS = 0$ so that $\mathrm{ann}_L$ and $\mathrm{ann}_R$ set up an (antitone) Galois connection between the posets of subspaces of $X$ and $X^*$. Various things then follow by abstract nonsense. For instance, $\mathrm{ann}_R \circ \mathrm{ann}_L$ and $\mathrm{ann}_L \circ \mathrm{ann}_R$ are abstract closure operators and the associated "closed subspaces" of $X$ and $X^*$ are put into order-reversing bijection by $\mathrm{ann}_L$ and $\mathrm{ann}_R$. In light of the fact that the range of $\mathrm{ann}_L$ consists of weak-star closed subspaces and the range of $\mathrm{ann}_R$ consists of weakly closed subspaces, it is natural to wonder whether these closure operators are, in fact, equal to the weak and weak-star closure operators. In essence, I am asking the following.
Question 1 (answered): Let $S$ be a subspace of $X$ and let $x \in X$. If $\varphi S = 0$ implies $\varphi(x) =0$ for all $\varphi \in X^*$, does it follow that $x$ is in the weak closure of $S$?
Question 2: Let $S$ be a subspace of $X^*$ and let $\varphi \in X^*$. If $S x= 0$ implies $\varphi(x) =0$ for all $x \in X$, does it follow that $\varphi$ is in the weak-star closure of $S$?
Thank you in advance for any answers or clarification on surrounding issues.
Added: I've managed to answer Question 1 affirmatively. First recall that the weak closure of $S$ is the same as the norm closure of $S$. More generally, the weak closure and norm closure coincide for convex subsets of $X$ (Conway, A Course in Functional Analysis, Theorem V.1.4). The Hahn-Banach Theorem implies the following statement: If $S \subset X$ is a subspace and $x \in X$ is a positive distance $d$ away from $S$, then there exists a functional $\varphi \in X^*$ with $\|\varphi\| = 1$ and $\varphi(x) = d$ and $\varphi S = 0$. It is easy to answer Question 1 using these two facts.