Number of elements of order $n$: explain why $\Bbb{Z}_4\times\Bbb{Z}_4$ and $\Bbb{Z}_4\times\Bbb{Z}_2\times\Bbb{Z}_2$ are not isomorphic.
The fact that $\Bbb Z/4\Bbb Z\times\Bbb Z/4\Bbb Z\not\cong\Bbb Z/4\Bbb Z\times\Bbb Z/2\Bbb Z\times\Bbb Z/2\Bbb Z$ is implied by $\Bbb Z/4\Bbb Z\not\cong\Bbb Z/2\Bbb Z\times\Bbb Z/2\Bbb Z$ (which is much easier to check: one of these groups has no element of order $>2$ while the other does) and related to the Fundamental Theorem of finite(ly generated) Abelian groups. Long story short: one can decompose (non-cyclic) abelian groups into direct products of cyclic groups of prime power order. However, splitting prime powers in this process produces non-isomorphic abelian groups.
As I am aware, this might be a bit over the top for this simple question. A simpler thing to observe is the following: $$\operatorname{ord}((a_1,\dots,a_n))=\operatorname{lcm}(\operatorname{ord}(a_1),\dots,\operatorname{ord}(a_n))$$ It is not hard to see that $\Bbb Z/4\Bbb Z$ has $2$ elements of order $4$, $1$ of order $2$ and $1$ of order $1$ while $\Bbb Z/2\Bbb Z$ has $1$ element of order $2$ and $1$ of order $1$. Combining these gives a much more efficient way of computing the number of elements of a given order.
The process always works in the sense that if you find different numbers of elements of a given order (dividing the group order) they cannot be isomorphic as isomorphisms preserve the order of elements. However, the converse is not necessarily true, that is there are groups of the same order with the same number of elements for every order which are not isomorphic (see here for a counterexample).
Let's consider the group $\Bbb Z/4\Bbb Z\times\Bbb Z/2\Bbb Z\times\Bbb Z/2\Bbb Z$. Write $\Bbb Z/4\Bbb Z=\{0,1,2,3\}$ and $\Bbb Z/2\Bbb Z=\{0,1\}$. As $\operatorname{lcm}(n_1,n_2,n_3)\ge\max\{n_1,n_2,n_3\}$ we conclude that for $(a_1,a_2,a_3)$ having order $2$ we have $a_1\ne1,3$. This leaves $2\cdot2\cdot2=8$ possibilites. However, if $a_1=a_2=a_3=0$ then $\operatorname{ord}(a_1,a_2,a_3)=1<2$. But this is the only exception to be found. Hence, there are exactly $7$ elements of order $2$.
For $\Bbb Z/4\Bbb Z\times\Bbb Z/4\Bbb Z$ we are immediately down to at most $4$ possibilites ($3$ remain in the end). This shows that they cannot be isomorphic.
It follows from the Fundamental Theorem of Finite(ly-generated) Abelian Groups. But this is a little glib on my part.
Yes, counting the number of elements of order two is sufficient. Trial & error will work since the groups are small enough.