Prove a group of order 12 must have an element of order 2

Hint: Here is a simple proof idea that every group of even order must have an element of order $2$.

Pair every element in $G \backslash \{ e \}$ with its inverse. If all pairs consist of two different elements then $G \backslash \{ e \}$ would have an even number of elements.

What does it mean that $a=a^{-1}$?

$a=a^{-1} \Leftrightarrow a^2=e$. And since $a \neq e$ we get that $ord(a)=2$.

Approach without Sylow's theorem: By what you've shown, all you need to do is discount the possibility that all group elements have order $3$ or $1$. The only element with order $1$ is the identity. What can you say about a group that consists of the identity, and $11$ elements of order $3$?

Hint: the elements of order $3$ can be partitioned into pairs $\{g,h\}$ s.t. $h=g^2,g=h^2$.

Hint:
Consider the Sylow $2$-subgroups of $G,$ which have order $4.$

Hope this helps.