Question: So I know all the orders of the elements in $(Z\_{12},+)$

$|[0]| = 1$

$|[1]| = 12$

$|[2]| = 6$

$|[3]| = 4$

$|[4]| = 3$

$|[5]| = 12$

$|[6]| = 2$

$|[7]| = 12$

$|[8]| = 3$

$|[9]| = 4$

$|[10]| = 6$

$|[11]| = 12$

I observe 2 things.
All the order of the elements divides |Z\_{12}|

And all the elements $x \in Z\_{12}$ which have $\gcd(x,12) = 1$ have order 12

Why is that true ?

Now if we consider $(U\_{12},\*)$ where $U\_{12} = \{[1],[5],[7],[11] \}$ we will see that all elements except the identity $[1]$ have order 2. Why is that true also ?

Answer: 
Let $G$ be a group .If $a\in G$ then $o(a)$ divides $o(G)$ (why?)
(Roll mouse here)

Here

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>  If $a\in G$ then consider $<a>$ then by Lagranges theorem,$o(<a>)$ divides $o(G)$ Now conclude..
> 
> 
> 

For the second part try to see that $a\in \mathbb Z\_{12}$ is a generator $\iff $ $\gcd(a,12)=1$ and since $[1]$ is a generator $o(a)=o([1])$
In the additive group $\mathbb{Z}\_n$, the order of an element $x$ is $n / \gcd(x, n)$.

To see why all the elements of the multiplicative group $\mathbb{Z}\_{12}^\times$ have order $2$, note that in general if $n = p\_1^{n\_1} \cdots p\_k^{n\_k}$ is the prime factorization of $n$, then the ring $\mathbb{Z}\_n$ is isomorphic to
$$\times\_{i = 1}^k \mathbb{Z}\_{p\_i}^{n\_i}$$
by the Chinese remainder theorem. For $p = 2$, the multiplicative group $\mathbb{Z}\_{2^e}^{\times}$ is isomorphic $\mathbb{Z}\_2 \times \mathbb{Z}\_{2^{e - 2}}$. For $p \geq 3$, $\mathbb{Z}\_{p^e}^{\times}$ is isomorphic to the cyclic group $\mathbb{Z}\_{p^{e - 1} (p - 1)}$.

In your case, we see that $\mathbb{Z}\_{12}^{\times} \cong \mathbb{Z}\_2 \times \mathbb{Z}\_2$, which explains why every nontrivial element has order $2$.