Using the pumping lemma for a specific language
Option 1: the word is in the language and it is of length $\geq n$, yet there is no guarantee that pumping it yields a contradiction. That is, the word can be used in the pumping lemma, but its not a good choice for proving non-regularity. Indeed, if $\text{LONGERB}$ is regular, then we know that there is a partition of $abab^{n+1}$ to three words, $xyz = abab^{n+1}$, where $|xy| \leq n$, $|y|&gt; 0$, and $xy^iz\in \text{LONGERB}$ for every $i\geq 0$. Thus, if for example $y = ba$, then $xy^iz$ is indeed in $\text{LONGERB}$ for every $i\geq 0$ -  so we cannot reach a contradiction to the assumption that $\text{LONGERB}$ is regular.

Option 2: the word is in the language and it is of length $\geq n$. Clearly it is good for pumping in the sense that it can be pumped for some $i$ so that $xy^i z \notin \text{LONGERB}$ - this is an easy exercise and the details are left to you.

Option 3: the word is of constant length that does not depend on the pumping constant $n$, so there is no guarantee that it is of length $\geq n$, so the pumping lemma says nothing about this word. Thus, we cannot use it to apply the pumping lemma on any language with a pumping constant $n$.

Option 4: you're right, the word cannot be used when applying the pumping lemma directly on the language $\text{LONGERB}$. However, it can be used to prove the non-regularity of $\text{LONGERB}$ indirectly by applying the pumping lemma on the language $\overline{\text{LONGERB}}$ which is the language of the words over $\{a, b\}$ where the longest substring containing only $b$’s is equal or shorter than the longest substring containing only $a$’s. Indeed,  assume by contradiction that $\overline{\text{LONGERB}}$ is regular. If you consider the word $w = a^n b^n$, it is longer that $n$ and it is in the language $\overline{\text{LONGERB}}$. Then, there is a partition $xyz = a^nb^n$, where $|xy| \leq n$, $|y|&gt; 0$, and $xy^iz\in \overline{\text{LONGERB}}$ for every $i\geq 0$. If you pump with $i = 0$, you get a contradiction. (BTW, I implicitly assumed that both a language and its complement have the same pumping constant $n$. This assumption is okay. Can you tell why?)