Edward from Aquinas College in England explained why each pair of statements fit around a $\Rightarrow$ or a $\iff$:
 

If Rover is a dog, he isn't a cat. If Rover's not a cat there are a lot of other things he could be.

 

If $2n-m<0$ then $m$ is bigger than $2n.$ It can't be the other way round: if $n\lt m,$ we don't know that $2n\lt m$

Sometimes, an example is helpful to show that something is not always true. It is not true that $n-m$ is not zero $\Rightarrow n+m$ is odd, because if $m=4$ and $n=2$ then $n-m$ is not zero, but $n+m$ is not odd.

$n$ is even, $n+1$ is odd.  $even+odd=odd.$ It is reversible as $odd-odd=even$

 

If $n$ is bigger than $m,$ then $n-m$ will be bigger than $0$ as $n\gt m$ rearranges to $n-m\gt0$

 

As $n$ is a positive integer, if $n^2$ is odd then $n$ must be odd.
Similarly, if n is odd then $n^2$ is odd. As $odd\times odd=odd$

 

If $n\gt2$ then $n^3$ will be greater than $5n$ as $n$ has to be an integer
The nearest integer after $2$ is $3$
$3^3=27$ which is greater than $5\times3=15$
It is reversible as if $n^3\gt5n,$ $n$ will be bigger than $2$


James from KEGS in England answered the question of whether there is more than one possible solution:
To begin with, I established all possible implicative relationships using the following visual aid:
 

There are 4 pairs of statements which imply the other, therefore these must be the 4 pairs in the solution, indicated by  

Rover is a dog $\Rightarrow$ Rover is not a cat, there are no other connections to these two. 

In the same way, $2n-m\lt0\Rightarrow n\lt m$

As with $n+m$ is odd $\Rightarrow n-m$ is not zero

This leaves the statements $n\gt1 \Rightarrow n\gt0$