Answer
The domain is the set of ordered pairs $\{(x,y)|x^2+4y^2\leq 4\}$.
The range is the set (a segment or the real line) $[0,2]$.
Work Step by Step
1) Domain
The function is given by $f(x,y)=\sqrt{4-x^2-4y^2}$. We have the restriction that the expression under the square root must not be negative which means that
$$4-x^2-4y^2\geq0\Rightarrow x^2+4y^2\leq 4.$$
So the domain will be the set of all ordered pairs $(x,y)$ such that $x^2+4y^2\leq 4$. We can write this set as $\{(x,y),x^2+4y^2\geq 4\}$.
2) Range
We can rewrite the function as
$$f(x,y) = \sqrt{4-x^2-4y^2} = \sqrt{4-(x^2+4y^2)}.$$
The sum $x^2+4y^2$ is never negative. When this sum is maximal the value of $f$ is minimal while when this sum is minimal the value of $f$ is maximal (since this sum is subtracted). The minimal value of $x^2+4x^2$ is $0$ and it is achieved when $x=0$ and $y=0$ so the maximal value for $f$ is $\sqrt{4-0} = 2$. The maximal value of the sum $x^2+4x^2$ is $4$ which we have shown when we calculated the domain so the minimal value of $f$ is $\sqrt{4-4}=0.$
This means that the range is
$$f(x,y)\in[0,2].$$
