We get $\text{Var}[Y] = \text{Var}[Z]$ from the symmetry in the setup: if you reverse the interval $[0,1]$, the distribution of $X_1$ and $X_2$ is unchanged, but $\min$ and $\max$ are swapped. Formally:
\begin{align}
\text{Var}[Y] &= \text{Var}[1-Y] \\
&= \text{Var}[1-\min\{X_1, X_2\}] \\
&= \text{Var}[\max\{1 - X_1, 1-X_2\}] \\
&= \text{Var}[\max\{X_1, X_2\}] \\
&= \text{Var}[Z].
\end{align}
The part that relies on distributions is the step where we replace $\max\{1-X_1, 1-X_2\}$ by $\max\{X_1, X_2\}$: that's because when $X_1, X_2$ are independent $\text{Uniform}(0,1)$, so are $1-X_1$ and $1-X_2$.
Looking at this argument, we see that it continues to work for $X_1, X_2 \sim \text{Uniform}(-1,2)$: in fact, the same transformation $t \to 1-t$ reverses the interval $[-1,2]$. More generally, we can do this for any symmetric distribution (where $X$ and $c-X$ are identically distributed for some $c$).
The other two facts are even more general: all they need is that addition and multiplication are commutative. Therefore $Y+Z = \min\{X_1, X_2\} + \max\{X_1, X_2\} = X_1 + X_2$: sorting $X_1$ and $X_2$ before adding them doesn't do anything. We conclude that
$$
\mathbb E[Y] + \mathbb E[Z] = \mathbb E[Y+Z] = \mathbb E[X_1 + X_2] = 2 \mathbb E[X_1] = 1.
$$
The value is unchanged, because $\frac{0+1}{2} = \frac{-1+2}{2} = \frac12$ is the value of $\mathbb E[X_1]$ in both problems.
Similarly, $\mathbb E[YZ] = \mathbb E[X_1X_2] = \mathbb E[X_1]^2 = \frac14$ in both cases.
Best Answer
Unsolicited advice: I feel it is better/clearer writing style to write things out in plain words.
I think $U(S)$ or $\text{Uniform}(S)$ are well-understood for finite sets $S$, but I would still write "where $U(S)$ denotes the uniform distribution over the set $S$" nonetheless (after the first instance of the notation) to eliminate any ambiguity.
I would use the notation $X_1, \ldots, X_n \overset{\text{i.i.d.}}{\sim} U(S)$ to denote i.i.d. samples from the aforementioned uniform distribution (do not use $U(S)^n$).