Too much long line

Let me redefine $L_2$ as a product of $\omega_1$ and the long line with lexicographical order, which seems more interesting. Note that $L_2$ is isomorphic to $\omega_1^2\times[0,1)$.

We can prove that both spaces are dense complete linear ordered set under the natural order. Completeness seems not trivial so I should sketch a proof of completeness of $I=\mu\times[0,1)$ for a limit ordinal $\mu$.

Divide $I$ into sets $I_{\alpha}=\{ \langle \alpha, r\rangle :r\in [0,1)\}$. Consider a subset $A$ of $I$ which is bounded above, so $A$ is contained in some $[(0,0), (\nu,0)]$ for some $\nu<\mu$.

Consider $B=\{\alpha\le\nu: I_\alpha\cap A\neq\varnothing\}$ and take $\beta=\sup B$. If $\beta\in B$, then finding a supremum of $A$ is reduced to finding a supremum of a set of $[0,1)$. If not, the point $(\beta,0)$ will be a supremum of $B$. (Added in Feb 10, 2019: we need to divide cases once more: because $\beta\in B$ does not guarantee $I_\beta$ has a supremum. If $r:=\sup I_\beta\in [0,1)$, $\langle\beta, r\rangle $ is a supremum of $A$. If $r=1$, $\langle\beta+1,0\rangle $ would be.)

Hence every connected subset of $I$ is an interval (see §24 of Munkres' Topology.) We can see that every compact interval in $L_1$ is separable. However $L_2$ is not.

We are dealing with $S\times \mathbb{R}$ with the topology given by the base $\{a \times I : a \in S, I$ : open in $\mathbb{R}\}$. I claim that such $S_{1} \times \mathbb{R}$ and another $S_{2} \times \mathbb{R}$ if and only if $|S_{1}| = |S_{2}|$.

Indeed, consider a (set-theoretical) bjection $h : S_{1} \rightarrow S_{2}$, and gather all the homeomorephisms between $a \times \mathbb{R}$ and $h(a) \times \mathbb{R}$ for $a \in S_{1}$ in order to construct entire homeomorphism.

Definitely, it may disturb the "order" of the components, but it does not matter when you focus on topological properties. If one focuses on the equivalence by order-preserving maps, it becomes entirely different story.

Edit: This answer doesn't apply to OP's question, which uses a nonstandard definition of long line. I had glossed over the definition assuming they would be equivalent. Mea culpa.

I think the answer to 1) is no. The largest ordinal you can embed into the long line is $\omega_1$ as per Asaf Karagila's answer in What uncountable ordinals live in the long line? , but you can easily embed larger ordinals in your long long line. Extending this line of thinking, we have a positive answer to 3).