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
Your interpretations for 1 and 2 are correct (except that $f : A \to \text{codomain}(f)$, not $f : \tau \to \text{codomain}(f)$). But they hide a very important part of the process, and from what you write I cannot tell whether you are aware of this.
A topology $\tau$ must satisfy some axioms: (1) the intersection of any two elements of $\tau$ is an element of $\tau$; (2) the union of any collection of elements of $\tau$ is an element of $\tau$; (3) the entire set, and the empty set, are elements of $\tau$.
Your version of 1, for example, says that $\tau$ is the set of all subsets of $A$ the form $f^{-1}(U)$ where $U$ is an element of the topology of $\text{codomain}(f)$.
This begs a big question: Does $\tau$ satisfy the axioms it is supposed to satisfy?
The answer is: Yes it does, but that requires some proof.
The outline of the proof is: Each of the three axioms (1), (2), and (3) for the topology on $\text{codomain}(f)$ implies the same axiom for $\tau$ itself.
Your interpretation for 3 has an error, the correct statement would be that $B$ is weakly compact if it is compact with respect to the subspace topology on $B$ that is induced from $\tau$.
Added in response to a comment from the OP: I don't know how to express compactness of $B$ directly in terms of $\tau$ other than to repeat the standard definition: for every collection of elements $\{U_i\}_{i \in I}$ of $\tau$ such that $B \subset \cup_i U_i$, there exists a finite subset $\{i_1,...,i_N\} \subset I$ such that $B \subset \cup_{n=1}^N U_{i_n}$. This does not imply that $B \in \tau$. Think about the compact subset $[0,1] \subset \mathbb R$ which is not an open subset of $\mathbb R$ (i.e. is not an element of the usual topology on $\mathbb R$).