Q1: If , you've got , which is undefined. Any other value of x gives you a defined output. The logic is the same for (b); if either term in the denominator is 0, the product is 0, so you get again.
Q2: must be 0 or larger, since you're squaring the value at the end (a negative value inside the parenthesis would become positive after squaring it).
The least that term can be is therefore 0, when . In that case . Any other value for x gives you a value larger than 2 (try plugging in a few values if you don't see that).
Q3: This is same logic as Q2. The term which is squared can only ever be 0 (if for (a) or for (b)) or larger. If you set it to 0, the function evaluates to 2 (for a) or -1 (for b).