Can I Systematically Determine the Cardinality of a Finite Set?

It's really dependent on how you built the set. The thing that can make things rather tricky is when things in the set are equal only one thing is counted. So usually inequalities are easy but the exact number you'd have to eleminate all equivalent elements.

For example, let $A$ be a set $\ A=\{1,2,3,4,5,6\}$ and $B$ be a set defined by: $$B=\biggl\{\frac{a-b}{a+b} : a,b\in A\biggl\}$$

Since $|A|=6$ and you are taking two elements in $A$ you get the easy inequality $|B| \le 6^2=36$.

Since $B$ has at least $1$ element $1 \le |B|$

Enumerateing through the values and evaluating them in the function $\frac{a-b}{a+b}$ i get the following table

$\begin{array}{rrrrrr} 0 & \frac{1}{3} & \frac{1}{2} & \frac{3}{5} & \frac{2}{3} & \frac{5}{7} \\ -\frac{1}{3} & 0 & \frac{1}{5} & \frac{1}{3} & \frac{3}{7} & \frac{1}{2} \\ -\frac{1}{2} & -\frac{1}{5} & 0 & \frac{1}{7} & \frac{1}{4} & \frac{1}{3} \\ -\frac{3}{5} & -\frac{1}{3} & -\frac{1}{7} & 0 & \frac{1}{9} & \frac{1}{5} \\ -\frac{2}{3} & -\frac{3}{7} & -\frac{1}{4} & -\frac{1}{9} & 0 & \frac{1}{11} \\ -\frac{5}{7} & -\frac{1}{2} & -\frac{1}{3} & -\frac{1}{5} & -\frac{1}{11} & 0 \end{array}$

So sorting out all the duplicates:

$$B=\left\{0, \pm\frac{1}{2}, \pm\frac{1}{3}, \pm\frac{2}{3}, \pm\frac{1}{4}, \pm\frac{1}{5}, \pm\frac{3}{5}, \pm\frac{1}{7}, \pm\frac{3}{7}, \pm\frac{5}{7}, \pm\frac{1}{9}, \pm\frac{1}{11}\right\}$$

So it looks like in your example $|B|=23$