$\newcommand{\N}{\mathbb{N}}$ $\newcommand{\F}{\mathbb{F}}$ $\newcommand{\coloneqq}{:=}$
It is my $\limsup$ and $\liminf$ sor a field $\F$:
$$\limsup(\liminf)x_{n} \coloneqq \inf\{x \in \F : x_{n} \leq (\geq) x \text{ for almost all $x_{n}$ }\}$$
Find a sequence $(a_{n})_{n \in \N}$ such that:
$$\inf\left(\{ a_{n} \,:\, n \in \N\}\right) < \liminf\left(\{ a_{n} \,:\, n \in \N\}\right) < \limsup\left(\{ a_{n} \,:\, n \in \N\}\right) < \sup\left(\{ a_{n} \,:\, n \in \N\}\right)$$
In its case, must the set $\{ a_{n} \,:\, n \in \N\}$ have a maximum?
I think is true but I do not whatr can I do...
Suppse by contraction it do not have a maximum. Call $S =\sup\{a_{n} : n \in \N\}$So $a_{n}$. By definition for any $n \in \N$ $a_{n} \leq S$
$a_1=-2$, $a_2=2$ and $a_n=(-1)^n$ for $n \ge 3$.
Regarding your second question. Denote $L=\limsup_n \{a_n\} \lt \sup_n \{a_n\}=S$. $L \lt S$ implies that it exists $N \in \mathbb N$ such that $a_n \le \frac{L+S}{2}$ for $n \ge N$. Therefore
$$\sup \{a_n \mid n \in \mathbb N\} = \sup \{a_n \mid n \lt N\}$$ and the $\sup$ of a finite set is attained.