Pushdown Automaton for $L = \{ w_1 w_2 : |w_1| =|w_2| , w_1 \neq w_2 \} $
The three languages you ask about are not very interesting:


$L_1$ is famously not context-free; it's usually written as $\{ ww \mid w \in \Sigma^* \}$. Note that the condition $|w_1| = |w_2|$ is obsolete. Standard entry-level proof techniques work.
$L_2 = \Sigma^+$ since every word $w$ (except $\varepsilon$) can be split into $w_1 = \varepsilon$ and $w_2 = w$; hence it's regular.
$L_3 = \emptyset$ since the two conditions are contradictory; hence it's regular.


As you see, neither gives you any inside into $L$. The languages discussed here are more helpful in this regard.



As for an NPDA for $L$, the answer you link already mentions that there are algorithms to get you from a grammar to an automaton; check a proof that NPDAs and CFGs are equally powerful.

That said, an idea for constructing an automaton from scratch is the same as for the grammar, even though it's a little messier to implement. Let $w$ be the whole input word, and rename $(w_1,w_2)$ to $(u,v)$ to avoid syntax overload. Then:


Note that if $u \neq v$ with $|u| = |v|$, there must be $i$ so that $u_i \neq v_i$. And, importantly, if there is no such $i$ if $u=v$.
Non-deterministically guess two input indices $j, l$  to compare.
Non-deterministicaly guess the middle index $k$ of the input.
Use states to verify that $w_j \neq w_l$.
Use states to verify that $|w|$ is even.
Use the stack to verify that $k = |w|/2$.
Use the stack to verify that $j = l - k$.
Reject if any of the verifications fail.


The main trick left is how to use the stack for the two checks at once; I'll leave that as an exercise.
Proving that the resulting automaton is instructive about the nature of non-determinism.