Understanding a boolean expression in λ-calculus
It's a very bad idea to omit parentheses. Correct is:
$$\lambda b . b (\lambda x y . y) (\lambda x y . x)$$
Next, it is confusing to reuse bound variables, so let me rename them:
$$\lambda b . b (\lambda x y . y) (\lambda u v . u)$$
And now we can apply this to $\lambda p q . p$ (which is just "true" with bound vraiables renamed yet again):
\begin{align*}
(\lambda b . b (\lambda x y . y) (\lambda u v . u)) (\lambda p q . p)
&amp;=_\beta (\lambda p q . p) (\lambda x y . y) (\lambda u v . u) \\
&amp;=_\beta (\lambda q . (\lambda x y . y)) (\lambda u v . u) \\
&amp;=_\beta (\lambda x y . y).
\end{align*}
So we did get false. In every step there was exactly one $\beta$-redex. By the way, writing $A B C$ in $\lambda$-calculus means $(A B) C$ by convention.

I will leave the other one as exercise.