Even without computing anything, one can guess at the onset that $(Y,Z)$ is probably not independent since $Y\leqslant Z$ almost surely. Now, the most direct way to compute the distribution of $(Y,Z)$ might be to note that, for every $(y,z)$,
$$
[y\lt Y,Z\leqslant z]=\bigcap_{i=1}^n[y\lt X_i\leqslant z],
$$
hence, for every $0\lt y\lt z\lt1$,
$$
P(y\lt Y,Z\leqslant z)=P(y\lt X_1\leqslant z)^n=(z-y)^n.
$$
In particular, using this for $y=0$ yields, for every $z$ in $(0,1)$,
$$
F_Z(z)=P(Z\leqslant z)=z^n.
$$
Thus, for every $(y,z)$ in $(0,1)$,
$$
F_{Y,Z}(y,z)=P(Y\leqslant y,Z\leqslant z)
$$
is also
$$
F_{Y,Z}(y,z)=P(Z\leqslant z)-P(y\lt Y,Z\leqslant z)=z^n-(z-y)^n\cdot\mathbf 1_{y\lt z}.
$$
One may find more convenient to describe the distribution of $(Y,Z)$ by its PDF $f_{Y,Z}$, obtained as
$$
f_{Y,Z}=\frac{\partial^2F_{Y,Z}}{\partial y\partial z}.
$$
In the present case, for every $n\geqslant2$,
$$
f_{Y,Z}(y,z)=n(n-1)(z-y)^{n-2}\mathbf 1_{0\lt y\lt z\lt1}.
$$
If $X_1$ and $X_2$ are independent, the first link you provided proves that $f(X_1)$ and $g(X_2)$ are independent. But that's not the situation that you have; here you're looking at $f(X_1, X_2)$ and $g(X_1, X_2)$.
In the case of max and min of independent uniform variables, the max and min are not independent, since their covariance is nonzero. Another way to see this: if you have knowledge of the value of the min, then the other variable (the max) cannot be less than this value; this constraint isn't present in the absence of that knowledge.
Best Answer
No, they are not. If we know that $L=x$, then the probability that $U<x$ is zero (while it is well possible that the prior probability was non-zero).