Because $A_k=\limsup\{X_n>a+\frac{1}{k}\}$ is increasing, it has a limit. It seems like $A_k\uparrow \limsup\{X_n>a\}$, but it’s wrong. There is a set $\{X_n=a+\frac{1}{n}\}$ belong to the right but don’t belong to any set of $\{A_k\}$.
So, what is $\lim_{k\to\infty}A_k$, and what is the gap between it and $\limsup\{X_n>a\}$
And I want to know the relation between $\{\limsup X_n>a\}, \limsup\{X_n>a\}, \{\limsup X_n\geq a\}$. I guess it is $\{\limsup X_n>a\}\subset \limsup\{X_n>a\} \subset \{\limsup X_n\geq a\}$, if it is right, how to prove? Thank you!
Since $A_k$ is an increasing sequence, your $\lim_{k\to\infty}A_k$ is $\bigcup_k A_k$. This is the event that "for some $k$, $X_n>a+1/k$ for infinitely many $n$", which is the same as "for some $\epsilon>0$, $X_n>a+\epsilon$ for infinitely many $n$". This is equivalent to the event $\limsup_{n\to\infty} X_n>a$.
Meanwhile "$\limsup\{X_n>a\}$" means "$X_n>a$ for infinitely many $n$". You are quite right: this implies that $\limsup X_n\geq a$, and is implied by $\limsup X_n > a$. Which of these implications is causing you difficulty? Whichever it is, maybe it will be helpful to apply the definition/fact that
$$\limsup b_n = \sup\{y: b_n > y \text{ for infinitely many }n\}.$$
For example, if $X_n>a$ for infinitely many $n$, then $a$ is a member of the set $\{y: X_n > y \text{ for infinitely many }n\}$, so $a\leq \sup\{y: X_n > y \text{ for infinitely many }n\}=\limsup X_n$.