If no two points on your manifold $M$ can be joined by both a spacelike curve and a timelike curve, I believe I can construct curves satisfying (1) and (2). However, I suspect that (3) can never also be satisfied. [Edit: I misread (3), and now I think it may be possible to satisfy (3) with a variant of the construction I gave for (1) and (2). See comments for details.] Here's a rough sketch of my argument. I've made the answer community wiki so that anyone can add clarifications or corrections.
As far as I can tell, the curves $A$, $B$, $G$, $K$, $Q$, and $V$ serve only to mark points on the other curves, $L$, $F$, $J$, $P$, and $U$. Every point in $M$ has an infinite number of timelike curves running through it, so any point in $M$ can be marked by a timelike curve distinct from $L, \ldots, U$. Therefore, we don't have to worry about $A, \ldots, V$ at all; we can just keep track of the points they mark, which I'll call
* $a = \mathcal{E}_{AL}$
* $b = \mathcal{E}_{BL}$
* $f = \mathcal{E}_{FG}$
* $j = \mathcal{E}_{JK}$
* $p = \mathcal{E}_{PQ}$
* $u = \mathcal{E}_{UV}$.
Starting with the curve $L$ and its distinct marked points $a$ and $b$, let's construct $F$, $J$, $P$, and $U$. Let $L'$ be a compact segment of $L$ containing $a$ and $b$ (and extending out beyond $a$ and $b$, for good measure). You should be able to construct a spacelike vector field $\xi$ on some open neighborhood of $L'$ which vanishes at $a$ but does not vanish anywhere else on $L'$. Every point on $L'$ has a neighborhood in which the flow $\phi$ generated by $\xi$ is well-defined for more than zero time. Since $L'$ is compact, it follows that $L'$ has a neighborhood in which the flow is well-defined for more than zero time.
Under the flow $\phi$, the curve $L'$ moves smoothly through $M$, and its tangent bundle $TL'$ moves smoothly through $TM$. 298
For every point $x$ on $L'$, the tangent space $T_xL'$ remains timelike for more than zero time during the flow. Thus, using comapctness, $L'$ remains timelike for more than zero time during the flow. Pick four times $t_F$, $t_J$, $t_P$, and $t_U$ at which $L'$ remains timelike, and call the images of $L'$ at these times $F$, $J$, $P$, and $U$.
Since $\xi$ vanishes at $a$, the curves $F, \ldots, U$ all pass through $a$. By tweaking $t_F, \ldots, t_U$, you should be able to ensure that $F, \ldots, U$ intersect $L$ only at $a$.
Let $f$ be the image of $b$ under the flow $\phi$ at time $t_F$. Define $j$, $p$, and $u$ similarly. Clearly $f$ lies on $F$, $j$ lies on $J$, and so forth. Notice that $b$, $f$, $j$, $p$, and $u$ lie on a spacelike curve: the path of $b$ under $\phi$. Therefore, by our initial assumption, no timelike curve can pass through two or more of these points.
We've now constructed curves with marked points satisfying (1) and (2).
Finally, here's why I suspect that (3) can never also be satisfied. [Edit: I misread (3), and now I think it may be possible to satisfy (3) with a variant of the construction I gave for (1) and (2). See comments for details.] Suppose we have some curves with marked points satisfying (1) and (2). I suspect that (2) prevents the light cones of $f$, $j$, $p$, and $u$ from covering all of $L'$. 490