The solution is using a few special properties of the exponential distribution.
Claim 1. If $A$ and $B$ are independent exponential random variables with parameters $\alpha$ and $\beta$, then, conditioned on $A < B$, the random variables $U := A \land B$ and $V := (A \lor B) - (A \land B)$ are independent exponential random variables with parameters $\alpha + \beta$ and $\beta$ respectively. (Similarly if conditioning on $B < A$ instead, the parameters would be $\alpha + \beta$ and $\alpha$ instead.)
The next claim is a corollary of the first. It is not necessary for solving your question but it is somewhat relevant given the author's rather casual tendency to drop the conditioning.
Claim 2. Consequently, the distribution of $U := A \land B$ (not conditioned on any event) is also exponential with parameter $\alpha + \beta$.
If we assume these claims, the (1) and (2) follow easily.
For (1): $E[A(A \land B) \mid A < B] = E[(A \land B)^2 \mid A < B]$ is direct. Then $E[(A \land B)^2 \mid A < B] = E[(A \land B)^2]$ follows from Claim 2. Finally, note that the second moment of an $\text{Exponential}(\alpha + \beta)$ distribution is $2/(\alpha + \beta)^2$.
(Actually, the step $E[(A \land B)^2 \mid A < B] = E[(A \land B)^2]$ is not really necessary, since by Claim 1 you already know that $A \land B$ conditioned on $A < B$ is exponential with parameter $\alpha + \beta$, but maybe the author had a different thought process.)
For (2): $A'$ is presumably defined as $V := (A \lor B) - (A \land B)$ (which is equivalent to $A-B$ when conditioned on $B < A$). So,
\begin{align}
&E[A(A\land B) \mid B<A]
\\
&=E[(U+V)U \mid B < A]
\\
&= E[U^2 \mid B < A] + E[VU \mid B < A]
\\
&= E[U^2 \mid B< A] + E[V \mid B<A] E[U\mid B,A]
& \text{$U$ and $V$ are cond. indep. given $B <A$}
\\
&= \frac{2}{(\alpha + \beta)^2} + \frac{1}{\alpha} \frac{1}{\alpha + \beta}.
\end{align}
Frankly I think the author's steps are a little misleading, but maybe I am missing a simpler way of approaching this.
Proof of Claim 1.
Let $U=A\land B$ and $V= A \lor B - A \land B$.
We have
\begin{align}
P(U > u, V > v, A < B)
&= P(A > u, B-A > v)
\\
&= \int_u^\infty \int_{a+v}^\infty f_A(a) f_B(b) \, db \, da
\\
&= \int_u^\infty \alpha e^{-\alpha a} \int_{a+v}^\infty \beta e^{-\beta b} \, db \, da
\\
&= e^{-\beta v} \int_u^\infty \alpha e^{-(\alpha + \beta) a} \, da
\\
&= \frac{\alpha}{\alpha + \beta} e^{-(\alpha + \beta)u} e^{-\beta v}.
\end{align}
Consequently, since you know that $P(A<B)=\frac{\alpha}{\alpha + \beta}$,
$$P(U>u, V>v \mid A < B) = e^{-(\alpha + \beta)u} e^{-\beta v}$$
which shows that, conditioned on $A < B$, the random variables $U$ and $V$ are independent exponential random variables with parameters $\alpha + \beta$ and $\beta$ respectively.
The proof when conditioning on $B<A$ is very similar.
For Claim 2, just note that
$$P(U > u) = P(U>u \mid A < B) P(A<B) + P(U>u \mid B<A)P(B<A)
= e^{-(\alpha + \beta)u} (P(A<B)+P(B<A))
= e^{-(\alpha + \beta)u}.$$
Alternatively, note that you can brute force compute these expressions directly without knowing about Claims 1 and 2.
\begin{align}
E[A(A\land B) \mathbf{1}_{A < B}]
&= \int_0^\infty \int_a^\infty a^2 f_A(a) f_B(b) \, db \, da
\\
&= \int_0^\infty a^2 \alpha e^{-\alpha a}\int_a^\infty \beta e^{-\beta b} \, db \, da
\\
&= \int_0^\infty a^2 \alpha e^{-(\alpha + \beta) a}
\\
&= \frac{\alpha}{\alpha + \beta} \frac{2}{(\alpha+ \beta)^2}
\end{align}
\begin{align}
E[A(A\land B) \mathbf{1}_{B < A}]
&= \int_0^\infty \int_b^\infty ab f_A(a) f_B(b) \, da \, db
\\
&= \int_0^\infty b\beta e^{-\beta b} \int_b^\infty a \alpha e^{-\alpha a} \, da \, db
\\
&\overset{*}{=} \int_0^\infty b\beta e^{-\beta b} \frac{1}{\alpha} (\alpha b + 1) e^{-\alpha b} \, db
\\
&= \frac{\beta}{\alpha(\alpha + \beta)} \int_0^\infty (\alpha b^2 + b) (\alpha + \beta) e^{-(\alpha + \beta)b} \, db
\\
&= \frac{\beta}{\alpha (\alpha + \beta)} \left(\frac{2\alpha}{(\alpha + \beta)^2} + \frac{1}{\alpha + \beta}\right)
\\
&= \frac{\beta}{\alpha + \beta} \left(\frac{2}{(\alpha + \beta)^2} + \frac{1}{\alpha(\alpha + \beta)} \right)
\end{align}
where the starred equality is from integration by parts:
$\int_b^\infty a \alpha e^{-\alpha a} \, da = \frac{1}{\alpha} \int_{\alpha b}^\infty u e^{-u} \, du = \frac{1}{\alpha} e^{-\alpha b} (\alpha b+1)$.
