An explanation for the result that $X$, $1-Y$, and $Y-X$ have the same distribution in this case is as follows.
First, consider a plane with coordinate axes $u$ and $v$ and let
$(U,V)$ be a random point in the plane chosen according to some
joint density function $f_{U,V}(u,v)$. $U$ and $V$ need not be
independent random variables. Then,
$$\begin{align*}
P\{X > \alpha\} &= P\{\min(U,V) > \alpha\} = P\{U > \alpha, V > \alpha\},\\
P\{1-Y > \alpha\} &= P\{1-\max(U,V) > \alpha\} = P\{\max(U,V) < 1 - \alpha\}\\
&= P\{U < 1- \alpha, V < 1- \alpha\},\\
P\{Y-X > \alpha\} &= P\{\max(U,V)-\min(U,V) > \alpha\}\\
&= P\{U-V > \alpha\} + P\{V-U > \alpha\}.
\end{align*}$$
These three probabilities can be found in the general case by
integrating $f_{U,V}(u,v)$ over the appropriate region which can be
described in the three cases respectively as
the northeast quadrant of the plane with southwest corner
$(\alpha, \alpha)$
the southwest quadrant of the plane with northeast corner
$(1-\alpha, 1-\alpha)$
the half-plane below the line $v < u - \alpha$ and the half-plane above the line $v > u + \alpha$
So much for generalities. If the random point $(U,V)$ is uniformly
distributed on a region $A$ of the plane (that is,
$f_{U,V}(u,v)$ is nonzero and constant for $(u,v) \in A$,
$f_{U,V}(u,v) = 0$ for $(u,v) \notin A$) and $B$ is any
region of the plane, then
$$P\{(U,V) \in B\} = P\{(U,V) \in A\cap B\}
= \frac{\mathrm{Area}(A\cap B)}{\mathrm{Area}(A)}.$$
In particular, if we can compute areas via mensuration
formulas learned in school, we do not need to integrate
formally.
Finally, in the special case when $A$ is the unit-area square with
opposite corners $(0,0)$ and $(1,1)$, and $\alpha$ is a number between
$0$ and $1$,
$$\begin{align*}
P\{X > \alpha\} &= P\{U > \alpha, V > \alpha\}\\
&= P\{(U,V) \in ~\mathrm{square~with~opposite~corners}~ (\alpha,\alpha)
~ \mathrm{and}~ (1,1)\\
&= (1-\alpha)^2,\\
P\{1-Y > \alpha\} &= P\{U < 1- \alpha, V < 1- \alpha\}\\
&= P\{(U,V) \in ~\mathrm{square~with~opposite~corners}~ (0,0)
~ \mathrm{and}(1-\alpha,1-\alpha)~ \\
&= (1-\alpha)^2,\\
P\{Y-X > \alpha\} &= P\{U-V > \alpha\} + P\{V-U > \alpha\}\\
&= P\{(U,V) \in ~\mathrm{triangle~with~corners}~ (\alpha,0),
(1,1-\alpha) ~\mathrm{and}~(1,0)\}\\
&\quad \quad
+ P\{(U,V) \in ~\mathrm{triangle~with~corners}~ (0,\alpha),
(1-\alpha,1) ~\mathrm{and}~(0,1)\}\\
&= \frac{1}{2}(1-\alpha)^2 + \frac{1}{2}(1-\alpha)^2 = (1-\alpha)^2.\\
\end{align*}$$
So the complementary cumulative distribution of the three
random variables $X$, $1-Y$ and $Y-X$ is the same $(1-\alpha)^2$
in this case,
and so the three random variables have the same density function
$2(1-\alpha)$, $0 \leq \alpha \leq 1$.
You will need to either
1) look at the bivariate distribution of $X$ and $Y$ in order to figure out what region of the pdf for $(X,Y)$ corresponds to $U$ and $W$, or
2) alternatively, make an algebraic argument in terms of the cdf - e.g.
$P(W\leq w)=P(X\leq w , Y\leq w)$ ...
That hint you mentioned doesn't help you any if you don't know what you're supposed to do with the standard uniforms.
Best Answer
Obviously $XY=UV$ therefore the calculation of $\mathbb{E} XY$ is really easy !