You don't find $w$. $w$ is a "given", in the following sense. For the weak formulation of the problem to make sense, the statement
$$ u|_{\partial\Omega} = g $$
is in fact the following statement: $\exists$ a fixed $w \in H^1(\Omega)$ such that the trace of $w$ to $\partial \Omega$ is equal to the trace of $u$.
The relevant section in Evans is trying to explain this. Basically what he is trying to say is that the intuition for the Dirichlet problem with "strong" solutions, where you prescribe boundary value as some continuous function $g$ on the boundary, must be replaced by an appropriate weak version defined relative to the trace operator to hypersurfaces, when you consider the weak formulation of the problem. This is because a solution $u$, as an object in the space $W^{1,2} = H^1$, is only an equivalent class of functions defined up to sets of measure zero. If $\Omega$ is a sufficiently regular open set, $\partial\Omega$ has measure zero, so it is meaningless to state that $u$ coincides with $g$ on $\partial\Omega$, since $u$ can always be modified on just $\partial\Omega$ to give any value you want there.
You should compare this to, for example, Theorem 8.3 in Gilbarg and Trudinger, Elliptical partial differential equations of second order, which states
Let [$L$ be an elliptic operator]. Then for $\psi \in W^{1,2}(\Omega)$ and $g,f^i\in L^2(\Omega)$, $i = 1 , \ldots, n$, the generalized Dirichlet problem, $Lu = g + D_i f^i$ in $\Omega$, $u = \psi$ on $\partial\Omega$ is uniquely solvable.
First we look for a distributional solution. Remember that, as an distribution, $\Delta u$ is defined by $$\langle\Delta u,v \rangle=-\langle\nabla u,\nabla v\rangle,\ \forall\ v\in C_0^\infty(\Omega). \tag{1}$$
From $(1)$, we can say that a solution in the distributional sense, is a function $u\in H^1(\Omega)$ with $Tu=g$ satisfying $$\int_\Omega \nabla u\nabla v=0,\ \forall\ v\in C_0^\infty(\Omega). \tag{2}$$
By density we may conclude from $(2)$ that $$\int_\Omega \nabla u\nabla v=0,\ \forall v\in H_0^1(\Omega). \tag{3}$$
This is not the only weak formulation for this problem, however, it is the one which comes from a variational problem, to wit, let $F:\{u\in H^1(\Omega):\ Tu=g \}\to \mathbb{R}$ be defined by $$Fu=\frac{1}{2}\int_\Omega |\nabla u|^2.$$
Note that $(3)$ can be rewritten as $$\langle F'(u),v\rangle=0,\ \forall\ v\in H_0^1(\Omega). \tag{4}$$
Also note that, once $\{u\in H^1(\Omega):\ Tu=g \}$ is a closed convex set of $H^1(\Omega)$ and $F$ is a coercive, weakly lower semi continuous function, we have that $F$ has a unique global minimum which satisfies $(4)$.
By not considering any kind of derivative of $u$, you can also use another weak formulation: let $C_0^{1,\Delta}(\overline{\Omega})=\{u\in C_0^1(\overline{\Omega}):\ \Delta u \in L^\infty(\Omega)\}$. A "very" weak solution, is a function $u\in L^1(\Omega)$ satisfying $$\int_\Omega u\Delta v=-\int_{\partial\Omega}g\frac{\partial v}{\partial \nu },\ \forall\ v\in C_0^{1,\Delta}(\overline{\Omega}).$$
In your setting, I mean, when $g\in H^{1/2}(\Omega)$, it can be proved that both definitions are equivalent. For references, take a look in the paper Elliptic Equations Involving Measures from Veron. It has a PDF version here. Take a look in page 8.
To conclude, I would like to adress @JLA, which gave a comment in OP; in the end, what we really want is a $H^2$ function (or more regular), because we are working with the Laplacean and it is natural to have two derivatives.
It can be proved, by using regularity theory, that $u$ is in fact in $H^2$, however, there is a huge difference between proving that $u$ is in $H^2$ after finding it in $H^1$ by the above methods and finding directly $u\in H^2(\Omega)$ by another method. Note, for example, that none of the methods above, does apply if we change $H_0^1(\Omega)$ by $H^2(\Omega)$.
Best Answer
This is a question of regularity i.e. you know that a weak solution $u \in H^1_0(\Omega)$ exists so now you must prove that any weak solution of (I) is actually an element of $ C^2(\Omega) \cap C(\overline{\Omega})$. (Note: you didn't mention it but for $u$ to be a classical solution you also need $u \in C(\overline{\Omega})$ for the boundary condition to make sense).
Your PDE is elliptic and the regularity theory of elliptic PDE is very well understood. A classical reference is Partial Differential Equations by Lawrence C. Evans (see Chapter 6) or a more in depth reference is Elliptic Partial Differential Equations of Second Order by Gilbarg and Trudinger. In fact, just googling 'regularity theory elliptic PDE' will likely return hundreds of references.
To answer your second question about 'explicitly solving' (II) it is practically impossible. To put it into perspective if $u$ satisfies (II) with $c=0$, $f \in C^\infty(\Omega)$ and $\Omega$ with $C^\infty$ boundary then standard regularity theory implies that $u \in C^\infty(\overline{\Omega})$. Hence, $u$ satisfies $-\Delta u =f $ in $\Omega$ and $u=0$ on $\partial \Omega$. We know then that we can write $u$ in terms of a Green's function, but only in very very specific cases of $\Omega$ can we actually work out that Green's function. If $c \neq 0$ there are still 'Green's functions' but these are even more challenging to calculate than before! This is why mathematicians gave up on trying to 'solve' PDE a long time ago - it is much easier (and often more useful) to understand the underlying theory.