a) If we want the function to be continuous at x = 1, does that mean that both of the one-sided limits equal to 2? In that case, would I plug in that y-value and a random x-value that satisfies the restrictions to find the possible values of a and b?
b) x cannot be -2
x has to be bigger than or equal to -b.
$\lim\limits_{x \to 1^{-}} f(x)=\frac{3}{1+2}-a = \lim\limits_{x \to 1^{+}} f(x)=\sqrt{b+1}=2=f(1)$
$a=-1, b=3$