It seems as if you are misunderstanding the definition of the commutator. It is not the center of the group.
Many times denoted $[G,G]$, it is defined to be the subgroup generated by the set $\{xyx^{-1}y^{-1} : x, y \in G\}$, where I am using multiplicative notation.
In your example $\mathbb{Z}$, an element in the commutator will be, for example with $x = 5$ and $y = 7$: $5 + 7 - 5 - 7$. You should be able to answer PVAL's question above, and this should give an indication of how to solve the problem.
The homomorphism proof as it currently stands is incorrect.
You have:
\begin{align*}
\phi(n_1)+\phi(n_2) & \color{red}{=} \phi(n_1+n_2) \\
& \color{red}{=} \phi(n_1)+\phi(n_2) \\
& = 7n_1+7n_2 \\
& = 7(n_1+n_2)
\end{align*}
You have the important elements here, but they're arranged in a confusing way, that I would consider incorrect.
Specifically, the red equalities above are incorrect.
If you have a group $(G,+)$, and another group $(H,\oplus)$, a group homomorphism $\phi:G\to H$ is a function $\phi$ such that, for all $x,y\in G$:
$$\phi(x+y) = \phi(x)\oplus\phi(y)$$
Note that there are different operations "inside" $\phi$'s brackets vs outside of them.
This is because the elements $x,y\in G$ (so are combined using $G$'s operation), but $\phi(x),\phi(y)\in H$ (so are combined using $H$'s operation).
A homomorphism is a function that satisfies the above.
So, to prove something is a homomorphism, you need to prove that it always satisfies that equation.
Let $\phi:\mathbb Z\to\mathbb Z$ be defined by $x\mapsto 7x$.
Then, we have that $\phi(x+y) = 7(x+y)$.
We also have that $\phi(x)+\phi(y) = 7x+7y$.
Can we show that these expressions are equal? If we can, then $\phi$ is a homomorphism.
Showing they're equal isn't too bad, what you do is say that:
$$\phi(x+y) = 7(x+y) \color{red}{=} 7x+7y = \phi(x)+\phi(y)$$
Everything here I've written is just "writing what $\phi(x)$ means" besides the equality in red, which using distributivity of addition in $\mathbb Z$.
But, what we've done here is start with $\phi(x+y)$, and show how that must be equal to $\phi(x)+\phi(y)$, so $\phi$ must be a homomorphism.
Best Answer
Hint: If $\phi([1])=[8]$, then $\phi([2])=[8]+[8]=[4]$, $\phi([3])=[4]+[8]=[0]$, and so on. So, you have enough information to know an explicit description of $\phi$.