1. f1 and f4 are functions. f2 and f3 are not.
2. Letting A and B be the left and right sets. \$$\forall a \in A, f1(a) \in B \wedge f4(a) \in B \$$. However, for both f3 and f2 \$$\exists a \in A \text{ s.t. } f(a) \notin B\$$. For f2 it's \$$\circ\$$, for f3 it's \$$\bullet\$$.
3. To have an injective, but not surjective, function \$$|A| \lt |B|\$$. An example of this would be \$$A = \\{1,2,3\\}\$$ and \$$B = \\{1,2,3,4,5,6\\}\$$ with \$$f (a) = 2a\$$. for each b in B either there is a (singular) a in A which maps to it or no a maps to it (at most one).
4. To have a surjective, but not injective, function we require the opposite: \$$|A| \gt |B|\$$. Using the same two sets from part 3, we can define \$$g(b) = \lceil \frac{b}{2} \rceil\$$. Each of the 3 elements of A will be mapped to, but each by 2 elements of B.

If the sizes of A and B are equal, then we can find functions which are both surjective and injective or functions which are neither. However we cannot find functions which are one or the other.