Lambda term satisfying two equations using Bohm Trees
This task can be solved very well by observation and inserting simple functions.  

First, $\Delta$ should obviously be a function applying its argument to at least two terms (because it's $\lambda x y.$). So let's set $\Delta = \lambda f.fXY$, which gives $$X(Y(\lambda u.YuY)X)X=y_1
\\ 
X(Y(\lambda u.Yu(\lambda abc.cab))X)X = y_2$$
Now we see that $Y$ is mostly applied to other terms that seem complex enough to solve this task for themselves, so why not just set $Y=I(=\lambda t.t)$?
This leads us to $$X(XI)X=y_1
\\ 
X(X(\lambda abc.cab))X = y_2$$
$X$ is now an abstraction of at least two variables, but we see that the second one (it's always $X$ in the first application) can probably be ignored. Setting $X=\lambda p q.Mp$: $$M(\lambda q.MI)=y_1
\\ 
M(\lambda q.M(\lambda abc.cab)) = y_2$$
The same ways we did with $\Delta$, we see that M takes an argument with at most three abstractions, so we make $M=\lambda f.(f P Q R)$. Doing several beta reductions, this yields $$PQRQR=y_1
\\ 
RPQQR = y_2$$
with the obvious solutions $P=\lambda a b c d.y_1$ and $R=\lambda a b c d.y_2$ and the free parameter $Q$.
Now finally, we can insert these terms back into $M$, $X$ and $\Delta$ to yield the solution $$\Delta=\lambda f.f(\lambda p q.p(\lambda a b c d.y_1)I(\lambda a b c d.y_2))I$$ (setting $Q=I$ for brevity's sake).
There could very well be a smaller solution, but I think the derivation process is the simplest you can get with this one.