As @Peter Melech posted in the comments above, $P(U=1)=\frac{1}{36}$, since there is only one pair when $max(X_1,X_2)=1$ and it is when $X_1=1\; , X_2=1$.
On the other hand, there are 11 distinct combinations when $min(X_1,X_2)=1$ and it is true when $min(X_1=1,X_2=i)$, $1\leqslant i \leqslant 6$.
Hence $P(U=1,V=1)=\frac{1}{36}\, \frac{11}{36}$,
but $P(U=1 \cap V=1)=\frac{1}{36}$.
Therefore $U$ and $V$ are not independent, as $P(U=1)P(V=1)\neq P(U=1\cap V=1)$.
As @Peter Melech posted in the comments above,
$P(U=1)=\frac{1}{36}$, since there is only one pair when $max(X_1,X_2)=1$ and it is when $X_1=1\; , X_2=1$. On the other hand, there are 11 distinct combinations when $min(X_1,X_2)=1$ and it is true when $min(X_1=1,X_2=i)$, $1\leqslant i \leqslant 6$.
Hence $P(U=1,V=1)=\frac{1}{36}\, \frac{11}{36}$, but $P(U=1 \cap V=1)=\frac{1}{36}$.
Therefore $U$ and $V$ are not independent, as $P(U=1)P(V=1)\neq P(U=1\cap V=1)$.