If DFA has two states, which of the conditions hold?
You seem to be misunderstanding the statement of the exercise. It wants you to show that if $L$ is a language accepted by a DFA containing two states, then either $L$ is empty, or $L$ contains the empty word, or $L$ contains a word of length 1. Which of the cases holds depends on $L$. More explicitly:


If no state of the DFA is accepting, then $L = \emptyset$.
If the initial state of the DFA is accepting, then $\epsilon \in L$.
If the other state of the DFA is accepting, then either the state is reachable, in which case $\sigma \in L$ for some $\sigma \in \Sigma$, or else it is unreachable, in which case $L = \emptyset$.


More generally, the proof of the pumping lemma shows that if $L$ is accepted by a DFA containing $n$ states, then either $L = \emptyset$ or $L$ contains some word whose length is less than $n$. This is sharp, in the following sense: for each $n \geq 1$ and $m \in \{0,\ldots,n-1\}$, there is a language accepted by a DFA of length $n$ whose shortest word has length exactly $m$ (exercise).