$a^2-b^2 = x$ where $a,b,x$ are natural numbers Suppose that $a^2-b^2 =x$ where $a,b,x$ are natural numbers.
Suppose $x$ is fixed. If there is one $(a,b)$ found, can there be another $(a,b)$?
Also, would there be a way to know how many such $(a,b)$ exists?

You need to exploit the fact that the right hand side of your equation can be factored. For example for part (1) of the exercise, if $x$ is odd, say $x = 2n+1$ for some integer $n$, then
$$
x = 2n + 1 = y^2 - z^2 = (y - z)(y + z)
$$ 
Now try to consider a trivial factorization of $2n+1$ like $2n+1 = 1 \cdot (2n+1)$ and compare the two factorizations to get a system of equations
$$
\begin{align}
y - z &= 1\\
y + z &= 2n + 1
\end{align}
$$
I think you can take it from here, but feel free to ask if you get stuck.
