Multiplication table with identity and order 
This seems relatively easy, however I do not understand what is meant by the order of $x$ for each $x\in G\setminus\{3\}$ is equal to $2$. Can someone point me in the right direction? Clearly $G\setminus\{3\}=\{1,2,4\}$ but where can I go from here
 A: HINTS:
If the order of each $x$ (excluding $3$ which is the identity) is equal to $2$ then what do you get when you multiply $x$ with itself?

 you of course get the identity element which is $3$ in your case, this way you can fill the diagonal in your table

How to fill the rest of the table?
Since all grouptables a so called latin squares so you have to have each element once in each row and column, a sort of a sudoku.
You need to have a $2$ and a $4$ for somewhere in the first row

 $2$ cannot be in the first row second column since this would mean that $1$ is the identity, so the first row has elements $3,4,1,2$ of course in this order from left to right

The rest fills itself
Hope I could help
A: "order" mean how many times you must operate an element with itself to get the identity.  
$a^k = a*a*a*...... *a$.  If $a^k =e$ and $k$ is the smallest power so that is true, that is called the "order of $a$"[1].  In this group $a^2 = e =3$ always. (except maybe for $3$ where $3=3$.  But $3*3 = 3$ so.. yes, $3^2 = 3$....)
So you have
$\begin{array}*&|1&2&3&4\\
-&-&-&-&-\\
1&|3&?&1&?\\
2&|?&3&2&?\\
3&|1&2&3&4\\
4&|?&?&4&3\\
\end{array}$
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[1] One can prove for a finite group that $a^k = e$ must exist.  For an infinite group (such as $\mathbb Z, +$) it doesn't have to.  In that case we say something has infinite order.
In $\mathbb Z, +$ the number $1$ has infinite order because $1+1+1+.....$ will never equal $0$.
In $\mathbb Z_6, +$ then $1$ has order $6$ because $1+1+1+1+1+1 = 0$.  ANd $2$ has order $3$ because $2+2+2 = 0$.  And $3$ has order $2$ because $3+3 = 0$.  And $4$ has order $3$ because $4+4+4= 0$.  $5$ has order $6$ because $5+5+5+5+5+5 = 6$ and $0$ has order $1$ because $0 = 0$.
The identity $e$ always has order $1$.
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Post script.  Other than labeling there are $2$ groups with $4$ elements.  There are those in which one of the elements is such that $a * a =b \ne e$ and $a *a*a = c\ne e$ but $a*a*a*a = e$. (If you complete this you will get there is only one possible group.)  And the other is where all of the elements are that $a * a = e; b*b=e; c*c=e$.  If you complete this (as you are doing above) you will get that there is only one possible group.
