Let me emphasise when $X$ is a Banach space (or a normed space) and someone writes $X^{**}$, this is understood as the continuous linear functionals on $X^{*}$ with respect to the norm topology unless stated otherwise.
For your first question, the answer in general is no. For a counterexample take the Banach space $c_{0}$ which consists of all complex-valued sequences that converge to $0$. For every $n\in\mathbb{N}$ let $e_{n}\colon \mathbb{N} \to \mathbb{C}$ be given by $e_{n}(k) := \delta_{nk}$. Define $F\colon c_{0}^{*} \to \mathbb{C}$ by $F(f) := \sum_{n\in\mathbb{N}} f(e_{n})$. Note that this is well-defined since it can be shown that $(f(e_{n}))_{n\in\mathbb{N}} \in \ell_{1}$ and $\|f\|_{c_{0}^{*}} = \sum_{n\in\mathbb{N}} |f(e_{n})|$ for every $f\in c_{0}^{*}$. It quickly follows that $F \in c_{0}^{**}$. Now suppose for a contradiction there is some $x\in c_{0}$ such that $F(f) = f(x)$ for all $f\in c_{0}^{*}$. Then for each $k\in\mathbb{N}$ we can evaluate $F$ at the map $e_{k}^{*}\colon c_{0} \to \mathbb{C}$ given by $e_{k}^{*}(f) := f(e_{k})$, which is an element of $c_{0}^{*}$, to obtain that $x(k) = 1$. But then $x$ does not converge to $0$ and this contradicts that $x\in c_{0}$. Hence $F$ is a member of $c_{0}^{**}$ that is not given by a pointwise evaluation on $c_{0}$.
For your second question, note that if $X$ is a normed space (over $\mathbb{K}$) and $x\in X$, then you have that the map $f\mapsto f(x)$ from $X^{*}$ into $\mathbb{K}$ is linear. It is also continuous by the estimate $|f(x)| \leq \|f\|_{X^{*}} \|x\|_{X}$, so this map is a continuous linear functional on $X^{*}$ and is consequently an element of $X^{**}$. From this we obtain a map $J\colon X \to X^{**}$ such that each $x\in X$ induces a pointwise evaluation map $J(x)$ given by $[J(x)](f) = f(x)$ for all $f\in X^{*}$. This map $J$ is the canonical embedding of $X$ into $X^{**}$. As mentioned before, the canonical embedding $J$ associates each $x\in X$ with the corresponding pointwise evaluation which is an element of $X^{**}$.
Linking the evaluation map $J$ back to the first question, if $J$ is surjective then it does follow that every member of $X^{**}$ is a pointwise evaluation on $X$. Such normed spaces where the canonical embedding is surjective are called reflexive.
For your third question, one definition of the weak$^{*}$ topology on the dual $X^{*}$ of a normed space $X$ is to provide $X^{*}$ with the coarsest topology that makes each pointwise evaluation continuous. This is the same as the topology given by the neighbourhood base in the question you linked. The question now is whether this topology makes any other members of $X^{**}$ apart from pointwise evaluations continuous. This is not the case as can be seen, for example, here and here. Therefore the continuous linear functionals on $X^{*}$ with the weak$^{*}$ topology are precisely the pointwise evaluations on $X$. In particular, the map $F$ in the first question is an example of a linear functional that is norm continuous but not weak$^{*}$ continuous.
