Symmetric difference (SD) - suppose $A$ and $B$ are $2$ sets then symmetric difference of A and B is $(A-B)\cup(B-A) = (A\cup B)-(A\cap B).$
In question : U < V if the minimum element in the symmetric difference of the two sets is in U . Example: $\{1,2,3\} <\{2,3,4,5,6\}$
Symmetric difference is $\{1\} \cup \{4,5,6\}$.
Now Consider a smaller set. Suppose $S= \{1,2,3,4\}$
Now the given $2$ statements are about smallest and largest subset. So considering set $S$ and $\emptyset$ (empty set) will be helpful.
First take $U = \{1,2,3,4\}$ and $V = \{1,2\}$ (we can take any set other than ∅ and S)
$SD = \{3,4\}$ $($just exclude the elements which are common in the $2$ sets$)$
Minimum element of $SD$ is $3$ which is in $U$ and if we observe carefully minimum element will always be in $U.$ Whatever the $V$ is.
So acc. to the question $\{1,2,3,4\}$ is smaller than any other subset of $S.$ S2 is true.
Now consider
$U=\emptyset$ and $V= \{1,2\}$ (we can take any subset of S)
$SD = \{1,2\}$
The symmetric difference will always be equal to $V.$ So minimum element of $SD$ will always exist in $V$ when $U$ is $\emptyset.$
So acc. to the que, $\emptyset$ is greater than any other subset of $S.$ S1 is also true.
This is true even when $S= \{1,2,3,\ldots,2014\}.$
So answer is A. Both S1 and S2 are true