= Solution
For $0\leq u\leq1$, <independence> gives
$$
P(U>u)=P(X>u,Y>u)=(1-u)^2.
$$
Thus $U$ has <probability density function> $f_U(u)=2(1-u)$, and its <expected value> is
$$
\boxed{\mathbb EU=\int_0^1 2u(1-u)\,du=\frac13.}
$$
Pointwise, $U+V=X+Y$ and $UV=XY$. By linearity of <expectation>, $\mathbb EV=1-1/3=2/3$; by <independence>, $\mathbb E(UV)=\mathbb EX\,\mathbb EY=1/4$. Consequently the <covariance of two uniform order statistics> is
$$
\boxed{\operatorname{Cov}(U,V)=\mathbb E(UV)-\mathbb EU\,\mathbb EV
=\frac14-\frac29=\frac1{36}.}
$$
The <order statistics> are dependent despite the independence of the original observations.
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