What if x = 6?
The statement "If $x$ is a multiple of twelve then $x$ is a multiple of three."
is true if and only if it holds for every $x$
The statement "If $x$ is a multiple of three then $x$ is a multiple of twelve." does not hold for every $x$ so it is false.
Please stand corrected.
$P \rightarrow Q$, which is read as "If P then Q", is equivalent to $ (\neg P \lor Q)$, which is read as "either not P or Q". This statement is only false when the antecedent, P, is true while the consequent, Q, is false. Where P, Q are well formed formulas.
In this particular case you have taken the statement, If x is a multiple of 12, then x is a multiple of 3, and correctly converted it into its consequent, If x is a multiple of 3, then x is a multiple of 12. You now have to either prove the statement or find a counterexample (The question asks you to do this when it says "state whether the converse is true or false").
For example consider $x=9$. Substituting 9 for x in the converse of the original conditional statement we have: If 9 is a multiple of 3, then 9 is a multiple of 12.
Recall the definition of a multiple; In mathematics, a multiple is the product of any quantity and an integer. In other words, for the quantities a and b, we say that b is a multiple of a if $b = na$ for some integer n. (This is from wikipedia)
Evaluate the antecedent first, or "9 is a multiple of 3". Since $9=3^2=3*3$ we conclude that 9 is indeed a multiple of 3.
Evaluating the consequent second, which is "9 is a multiple of 12". We have it, from the definition of multiple that there exists $n \in \mathbb{N}$ such that $12n=9$. Since $n=\frac{9}{12}=\frac{3}{4}$ n is a rational number, which contradicts our assumption $n \in \mathbb{N}$. Hence we are entitled to conclude the negation of the consequent.
So we have a situation where the antecedent is true, while the consequent is false. Hence the conditional is false.