If $F$ is a family of linear functionals on a vector space $X$ then it induces a weak topology $w_F$ on $X$. The dual space of $(X,w_F)$ is the linear span of $F$ in the algebraic dual of $X$. In particular $(X,s)' = (X,w)'$ always holds.
This fully answers the first bullet. It also shows that there is no example as in the second bullet and that the procedure in the third bullet stabilizes at $(X,w)$ after one step.
The point is that if $\varphi \colon X \to \mathbb{F}$ is $w_F$-continuous then $U = \{x \in X \mid \lvert \varphi(x)\rvert \lt 1\}$ is a $w_F$-open neighborhood of $0$. By definition of the weak topology, there are $f_1,\dots,f_n \in F$ and $\varepsilon \gt 0$ such that $V = \bigcap_{i=1}^n \left\{x \in X \mid \lvert f_i(x)\rvert \lt \varepsilon\right\} \subseteq U$ and in particular $\bigcap_{i=1}^n \ker f_i \subseteq \ker \varphi $. Therefore $\varphi$ is a linear combination of the $f_i$ (a proof of this last assertion is here).
You can find a detailed proof in pretty much every book on topological vector spaces. A good reference is chapter 3 of Rudin's Functional Analysis, see in particular Lemma 3.9 and Theorem 3.10 (the Hausdorff assumption is only there because Rudin assumes locally convex spaces to be Hausdorff, as many authors do).
One goes about understanding a topology by looking at the basis, the "simplest" open sets of which all other sets are composed as unions. For a metric space, the basis consists of the balls, of course. And what's even nicer about topological vector spaces is that you only have to study what the basis sets look like at the origin, since topological vector spaces are shift-invariant.
The norm-topology's basis at the origin consists of the balls $ B(0,r) $. Easy enough. What do basis sets look like in the weak topology? Take a bounded linear functional $ f \in X^* $, and then look at $ \{ x \in X : |f(x)| < r \} $. You can hopefully imagine, this is "the space that is sandwiched between two hyperplanes". Now, this isn't quite the basis. To get a basis (at the origin), you need to consider all possible intersections of these "sandwiches". But the important thing is, you only get to "sandwich" in a finite number of directions $ f_1, \ldots, f_N $. So, in the infinite dimensional case, there's always some direction that you fail to sandwich. Hence, these basis sets are not bounded. This is counterintuitive, because in $ \mathbb{R}^3 $ (for example), you obtain a bounded set once you "sandwich" along the x-axis, y-axis, and z-axis (or in 3 directions of your choice).
Question 1 Let's first get clear on why the norm-topology contains the weak topology. Let $ U $ be open in the weak topology. Fix a point $ x_0 \in U $. We can find a basis set $ V $ with $ x \in V \subseteq U $. Recall, by "basis set", I mean a finite number of sandwiches, but we can be rigorous about this: for some numbers $ m_1 < M_1, \ldots m_N < M_N $ and $ f_1, \ldots f_N \in X^* $, the basis set $ V $ can be expressed
$$
V = \{ x \in X : m_1 < f_1(x) < M_1, \ldots, m_N < f_N(x) < M_N \}
$$
Now, it's pretty straightforward to find a ball $ B(x_0, r_0) $ that fits inside of $ V $. Do you see how? It relies on the boundedness of hte functionals $ f_1, \ldots f_N $. Thus we have
$$
x_0 \in B(x_0, r_0) \subseteq V \subseteq U
$$
which, you'll recall, shows that $ U $ is open in the norm topology (make sure you see why).
Question 1, Part b You then asked why $ x \mapsto |f(x)| $ is norm-continuous. But this is a composition of two continuous functions, the absolute value function and $ f \in X^* $ (which is continuous by definition).
Question 2 You asked about showing that $ g : x \mapsto \|Lx\| $ is weakly continuous, where $ L:X \rightarrow X $ is a bounded linear map. Surprisingly, it is not continuous with respect to the weak topology (in general). Indeed, pick $ X $ to be your favorite infinite-dimensional space, let's say $ X = \mathcal{l}^1(\mathbb{N}) $ and $ L = Identity : X \rightarrow X $. Then I claim $ g : x \mapsto \|Lx\| $ is not weakly continuous, since indeed $ g^{-1}((-a,a)) = B(0,a) $ is bounded, whereas weakly open sets in $ \mathcal{l}^1 $ are not bounded (!).
Addendum Fundamentally, the thing that's surprising here is that the norm $ x \mapsto \|x\| $ is not weakly continuous. To try to wrap our heads around this, consider the weak basis we discussed above, composed of finite intersections of "hyperplane sandwiches." This basis of the weak topology tells us that, in order for a function $ f : X \rightarrow \mathbb{R} $ to be weakly continuous, it is only allowed to change in a finite number of directions. The norm function, on the other hand, changes in all directions that you walk away from the origin.
Best Answer
That follows from a more general statement:
For $f_{0}$ a continuous linear functional with respect to $\tau$, then there is a neighborhood of $0$ in $X$ mapped by $f_{0}$ into $\{|x|<1\}$. Then there are neighborhoods $U_{1},...,U_{n}$ of $0$ in $X$ and $f_{1},...,f_{n}\in X'$ such that $f_{0}(f_{1}^{-1}(U_{1})\cap\cdots\cap f_{n}^{-1}(U_{n}))\subseteq\{|x|<1\}$.
For $x\in\ker(f_{1})\cap\cdots\cap\ker(f_{n})$, we have $|f_{0}(mx)|=m|f_{0}(x)|<1$ for all $m=1,2,...$, this forces $f_{0}(x)$ to be zero, and hence $x\in\ker(f)$.
A classic linear algebra fact states that $f=\displaystyle\sum_{i=1}^{n}c_{i}f_{i}$ for some constants $c_{i}$, and hence $f\in X'$.
Now every $f\in X'$ is certainly continuous with respect to $\tau$ by the very definition of induced topology $\tau$.
Now let $Q(X)=\{Q(x): x\in X\}$, where $\left<x^{\ast},Q(x)\right>=\left<x,x^{\ast}\right>$ for $x^{\ast}\in X'$ and $x\in X$. Then the weak$^{\ast}$ topology for $X^{\ast}$ is exactly the topology induced by $Q(X)$.