Equivalent conditions for weak and weak-$*$ convergence

Question

Say $E$ is a normed space over the field $\mathbb K$ ($\mathbb R$ or $\mathbb C$) and $E^{*}$ its dual space. The notations for weak and weak - $*$ convergence are $x_{n} \xrightarrow{w} x$ and $\phi_{n} \xrightarrow{w^{*}} \phi$ respectively.

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I have proven that:

1. If $E$ is Banach and $(\phi_{n})_{n} \subset E^{*}$ is such that $(\phi_{n}(x))_{n}$ converges in $\mathbb K$ for every $x \in E$ then there exists $\phi \in E^{*}$ with $\phi_{n} \xrightarrow{w^{*}} \phi$
2. If $E$ is reflexive and $(x_{n})_{n} \subset E$ is such that $(\phi(x_{n}))_{n}$ converges in $\mathbb K$ for every $\phi \in E^{*}$ then there exists $x \in E$ with $x_{n} \xrightarrow{w} x$
3. The proposition 2. is flase if $E$ is only assumed to be normed

My questions are:

Does 1. remain true if $E$ is only assummed to be normed?

Does 2. remain true if $E$ is only assummed to be Banach?

Counterexamples?

Answer 1

1. No (I assume that $E^*$ denoted the continuous dual space). Let $$E = \{x\in \ell_1: x\text{ has finite support}\}$$ equipped with norm $\|\cdot\|_1$. Let $\phi_n = \sum_{k=1}^n ke_k \in E^*$. Then for every $x\in E$, we have $\phi_n(x) \to \sum_{k=1}^\infty kx_k$ as $n\to\infty$. But $ \sum_{k=1}^{\infty} ke_k$ is an unbounded linear functional and thus does not belong to $E^*$.

2. David Mitra answered the question in comments.