Firstly, you don't seem to want to be doing us the favour of writing your question in a manner we'd like to see it!
Secondly, your notation is cumbersome for me to type it over here. So, set $G^{(n)}=G_n$ and similarly for $n-1$.
This is the way to go about it:
1) Note that identity element in $G$ belongs to $G_n$. If $g,h \in G_n$, then, there exists $x, y \in G$ such that $g=x^n$ and $h=y^n$. So, $gh^{-1}=x^ny^{-n}=(xy^{-1})^n$ Since, $gh^{-1}$ can be written in the form of $l^n$ for some $l \in G$,we have that $gh^{-1} \in G_n$. Hence, $G_n$ is a subgroup.
To prove it is normal, note that $n^{th}$ power map is a homomorphism from $G$ to $G$. Here's where you need to think, I don't want to kill the purpose of the home work, or so, you tell yourself, I don't know how to prove it!
For the next exercise, here's what I'll do:
Given, $\begin{align*}(ab)^n&=a^nb^n\\ab \cdot ab \cdots ab&=aa^{n-1}b^{n-1}b\\(ba)^{n-1}&=a^{n-1}b^{n-1}\\(ba)^n&=a^{n-1}b^{n}a\\b^na^n&=a^{n-1}b^{n}a\\b^na^{n-1}&=a^{n-1}b^n\end{align*}$
Make sure you justify each step above. I will elaborate if you seem to get lost somewhere.