= Solution
For independent <random variables>, $\{\min(X,Y)>u\}=\{X>u,Y>u\}$ and $\{\max(X,Y)\leq v\}=\{X\leq v,Y\leq v\}$. Factor their <probabilities> to obtain the <distribution functions of independent minima and maxima>:
$$
\boxed{F_U(u)=1-[1-F_X(u)][1-F_Y(u)],\qquad F_V(v)=F_X(v)F_Y(v)}.
$$
These formulas also allow atoms; no <probability density function> assumption is needed.
For independent unit-rate <exponential distributions>, $P(U>u)=e^{-2u}$ for $u\geq0$, so $U\sim\operatorname{Exp}(2)$. To establish the <independent minimum and gap of two exponential variables>, set $D=V-U$. On either ordering branch, the change of variables is $(X,Y)=(u,u+d)$ or $(u+d,u)$ with unit absolute <Jacobian determinant>. Both contribute $e^{-2u-d}$; ties have <probability> zero. Thus the <joint probability density> is
$$
f_{U,D}(u,d)=2e^{-2u-d}=(2e^{-2u})(e^{-d}),\qquad u,d>0.
$$
It factors into normalized exponential <probability density functions>, proving $U\sim\operatorname{Exp}(2)$, $D\sim\operatorname{Exp}(1)$ and their <independence>.
The maximum is $V=U+D$. A half-scaled unit-rate exponential has rate two, so the independent pair $(D,U)$ has the same joint law as $(X,Y/2)$. Consequently
$$
\boxed{V\overset d=X+\frac12Y}.
$$
Using the <expected values>, <variances> and <independence> of the summands,
$$
\boxed{\mathbb E V=\frac32,\qquad\operatorname{Var}(V)=\frac54}.
$$
The spacing factorization explains why an order statistic can be represented as a sum of independent waiting times.
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