Answer
$a.\qquad 1+\sqrt{3}$
$ b.\qquad \{(x,y)\quad|\quad x\in \mathbb{R}, \quad y\in[-2,2]\}$, see screenshot below
$ c.\qquad [1,3]$
Work Step by Step
$a.$
$F(3,1)=1+\sqrt{4-1^{2}}= 1+\sqrt{3}$
$b.$
Restriction because of the square root: $4-y^{2}\geq 0$
$4\geq y^{2}$
$|y|\leq 2\Rightarrow y\in[-2,2]$
Domain of F = $\{(x,y)\quad|\quad x\in \mathbb{R}, \quad y\in[-2,2]\}$
$c.$
Since $y\in[-2,2],$
$\sqrt{4-y^{2}}$ ranges from 0 to 2,
$0\leq\sqrt{4-y^{2}}\leq 2\qquad/+1$
$1\leq 1+\sqrt{4-y^{2}}\leq 3$
Range of F = $[1,3]$
