1. Show that if a is in G, where G is a finite group with identity e, then there exists n in positive integers such that a^n=e
2. Let G be a group and let a be one fixed element of G. Show that H (sub a)= {x in G: xa=ax}
Let $\displaystyle a \in G$ and say $\displaystyle n = |G|$ consider $\displaystyle a,a^2,...,a^n$. Now if $\displaystyle a^i = a^j $ for some $\displaystyle i>j$ then $\displaystyle a^{i-j} = e$ where $\displaystyle 1\leq i-j\leq n$. Otherwise if $\displaystyle a,a^2,...,a^n$ are all distinct then by pigeonholing it is a permutation of element $\displaystyle G$ and so $\displaystyle a^k = e$ for $\displaystyle 1\leq k\leq n$.
I assume you want to show $\displaystyle H_a$ is a subgroup of G. What have you done? Show what you tried to do.2. Let G be a group and let a be one fixed element of G. Show that H (sub a)= {x in G: xa=ax}
First, this is subgroup has a name. It is called the centralizer of $\displaystyle a$. And it is $\displaystyle H_a = \{ x\in G| xa = ax\}$. So basically it is the set of all $\displaystyle x$ that commute with all $\displaystyle a$. Let $\displaystyle x\in H_a$ then $\displaystyle ax = xa$ so $\displaystyle y(ax) = y(xa)$ but $\displaystyle y\in H_a$ so $\displaystyle ay=ya$ this means $\displaystyle a(yx) = (yx)a$ so $\displaystyle xy\in H_a$. This $\displaystyle H_a$ has the identity $\displaystyle e$. Now it has the associative property because G which contains it has that property. Finally if $\displaystyle x\in H_a$ then it means $\displaystyle ax = xa$ thus $\displaystyle axa^{-1} = xaa^{-1} \implies axa^{-1} = x$ thus $\displaystyle a^{-1} axa^{-1} = a^{-1} x\implies xa^{-1} = a^{-1} x$ and so $\displaystyle a^{-1} \in H_a$. So this proves it is a subgroup.