Don't try to do this in Java, just try to do it in pseudocode.
In the first question, you have two Lists of Integers; two lists of numbers. If there aren't any other rules given, we can assume that both lists are about n items long.
You're walking all the way through one of the lists, and doing something for each number in it. If there are n numbers, that's n steps. For each of those n steps, you're saying "does the other list have this number?"
To check and see if a list has a number in it, the worst case is that it doesn't, which means you have to check each item in the list to find out. That's also n steps, if the second list can be about as big as the first.
Putting those together:
Walking through a list of n items, and doing n things on each item.
Let me know what you think that comes up to, and I'll edit this answer after that.
For your second code snippet, split it up.
The first thing to happen is that you create a HashSet using all the items in the input. How long does it take to insert one item to a HashSet? If you insert n items, multiply the time for a single insert by n.
The second thing that happens is the new ArrayList is called. Checking the API to be sure of what it's doing:
http://docs.oracle.com/javase/7/docs/api/java/util/ArrayList.html#ArrayList(java.util.Collection)
It's creating an ArrayList using all of the items in the HashSet Collection. How long does it take to insert to an array list? Multiply that by the number of items (n) to get the answer.
Since these happen as separate steps, they're separate things. Whichever has a higher big-O complexity is the dominant one, and you can ignore the other. If they both have the same big-O complexity, that's the answer; you can ignore one, as O(2n) is the same complexity as O(n); constants (like 2x) get dropped.