The correct answer is 1 and 2:
L is not context-free.
L is not regular.
How to explain
If we have a language L and a string w where |w| >= p, and there exists a decomposition w = uvxyz such that |uv| <= p, |vy| > 0, and for all i >= 0, uvixy^iz is not in L, then we can conclude:
1. L is not context-free:
The pumping lemma for context-free languages states that for any string in a context-free language longer than the pumping length, we can "pump" a substring within the string and still get a valid string in the language. The existence of a decomposition where pumping results in a string not in the language contradicts this property.
2. L is not regular:
Regular languages are a subset of context-free languages, and any property true for context-free languages also applies to regular languages. Therefore, if L is not context-free, it cannot be regular.
3. None of the other choices is a valid conclusion about L:
Since we have already established that L is not context-free and not regular, the remaining options (L is regular or L is context-free) are not valid conclusions based on the given information.
Therefore, the correct answer is 1 and 2:
L is not context-free.
L is not regular.