Questions on limit superiors Given any bounded sequence $x(n)$, let $l$ be its limit superior. Let $\epsilon > 0$.


*

*Does there exist $n \in \mathbb{N}$ such that $x(n)<l+\epsilon$?

*Does there exist infinitely many $n$ such that $l-\epsilon < x(n)$?

 A: In fact, one of several equivalent definition which are commonly used for limit superior is that:
The number $l$ is called limit superior if


*

*for each $\varepsilon>0$ there exists $n_0$ such that for $n\ge n_0$ the inequality $x_n\le l+\varepsilon$ holds;

*for each $\varepsilon>0$ and for each $n_0$ there exists $n\ge n_0$ such that $x_n\ge l-\varepsilon$.


In symbols:
$$
(\forall\varepsilon>0)(\exists n_0)(\forall n\ge n_0) x_n\le l+\varepsilon\\
(\forall\varepsilon>0)(\forall n_0)(\exists n\ge n_0) x_n\ge l-\varepsilon
$$
So if this was definition which your book/course uses for limit superior, the properties follow directly from the definition.
If you are using another definition, you may try to prove the equivalence of two definitions of limit superior as an exercise. (BTW you might have mentioned what is your definition of $\limsup x_n$ in your post.) In the definition which I have mentioned above, the case that $\limsup x_n=+\infty$ must be dealt with separately.
See also these posts, where several commonly used definitions of limit superior are mentioned: 


*

*Two definitions of $\limsup$

*What is limit superior and limit inferior?

*Characterization of lim sup, lim inf

*Limit superior of a sequence is equal to the supremum of limit points of the sequence?

*Showing that two definitions of $\limsup$ are equivalent
A: 1) Yes. If to the contrary you have $x(n) \ge l + \epsilon$ for all $n$ then also you will have $l \ge l + \epsilon$, which can't happen. 
2) Yes. If to the contrary you have $l - \epsilon < x(n)$ for at most finitely many $n$, there is a point in the sequence beyond which $l - \epsilon \ge x(n)$ for all $n$. This would force $l - \epsilon \ge l$, which can't happen.
