A natural sample space is the collection of all "hands" of $5$ numbers. There are $\dbinom{20}{5}$ such hands, and they are all equally likely.
For b), there are $\dbinom{18}{5}$ ways to choose $5$ cards, none of which is $1$ or $2$. So our probability is $\frac{\binom{18}{5}}{\binom{20}{5}}.$ If desired, considerable simplification is possible.
For c), the simplest way is to subtract the answer to b) from $1$. This is because the event "at least one from $1$ or $2$" happens precisely if the event described in b) doesn't happen.
We can also solve the problem in a way close to the idea you mentioned in a comment. If we want a $1$ or a $2$ or both, there are three ways this can happen: (i) $1$ but not $2$; (ii) $2$ but not $1$; both $1$ and $2$.
The number of choices for (i) is $\dbinom{18}{4}$. The number of choices for (ii) is also $\dbinom{18}{2}$. And the number of choices for (iii) is $\dbinom{18}{3}$. Add up, and for the probability divide by $\dbinom{20}{5}$.