Is $(-2)^{\sqrt{2}}$ a real number? 
Is $(-2)^{\sqrt{2}}$ a real number?

Clarification: Is there some reason why $(-2)^{\sqrt{2}}$ is not a real number because it doesn't make sense why it shouldn't be a real number.
Mathematically we can define the value of $\sqrt{2}$ in terms of a limit of rationals. But the problem is is that some sequences will have values that are not defined for $(-2)^q$ where $q$ is a rational, such as $\dfrac{3}{2}$. Is this the reason why we can't define it or is there some other reason?

No. There are a countably infinite number of possible values of this expression. None of them are real. They are
$$2^{\sqrt{2}}(\cos (2k+1)\pi\sqrt{2} + i\sin(2k+1)\pi\sqrt{2})
$$
The principal value (taking $k=0$) is $2^{\sqrt{2}}(\cos\pi\sqrt{2} + i\sin\pi\sqrt{2})$.
