Is $(-2)^{\sqrt{2}}$ a real number?
I'll answer the general question of how we make sense of such expressions:
The value of such numbers is determined in one of two ways: by functional equations or by power series. A functional equation is any formula that relates the value of a function at two different points. Rules like $a^{b+c}=a^ba^c$ are examples of basic functional equations which can be rewritten as $f(a,b+c)=f(a,b)f(a,c)$ if that is preferred. An example of a function defined this way is $\Gamma(t):=\int_0^\infty x^{t-1}e^{-x}dx$. After applying integration by parts, we see that $t\Gamma(t)=\Gamma(t+1)$ holds, and we can use this to define $\Gamma(t)$ for any complex number that is not of the form $-n$ for $n\in\mathbb{N}$ (including $0$). This slightly weird way of expressing the undefined points is justified by the fact that at $-n$, $\Gamma(t)$ has a simple pole with residue $\frac{(-1)^n}{n!}$
Such values can also be determined by trying to make a power series make sense when values are plugged into it (for example, we might say $e^{i\pi}=-1$ because of what happens when we plug $i\pi$ into the power series for $e^x$). An example of a function defined this way is the matrix function $f(A)=e^A$. You can prove that the power series expression for $e^x$ around $0$ converges when interpreted as a statement about matrices. This function shares many of the properties of the complex function $f(z)=e^z$.
The wikipedia page on Analytic continuation has more info about extending (complex differentiable) functions.
We like these ways of doing things because they usually ensure that the properties we like the functions to have continue to hold. For example, extending via functional equation guarantees that the extended function also satisfies the functional equation, while power series are useful for preserving analytic properties of the function.
To deal with the question at hand now, your expression turns out to satisfy: $$(-2)^\sqrt{2}=2^{\sqrt{2}}(\cos((2k+1)\pi\sqrt{2}) + i\sin((2k+1)\pi\sqrt{2}))$$ which arises out of rewriting $a^x=\exp(x\log(a))$ and using the power series for $\exp(z)$. This is how we have chosen to define calculating generalized exponents. This is a multivalued expression, but for the principal value of $k=0$ we get $(-2)^\sqrt{2}=2^{\sqrt{2}}(\cos(\pi\sqrt{2}) + i\sin(\pi\sqrt{2}))$ which is not a real number because $\sin(\sqrt{2}\pi)\neq 0$.
By definition, $(-2)^\sqrt{2} = \exp(\sqrt{2} \log(-2))$. This is multivalued because $\log$ is multivalued: we have $\log(-2) = \log(2) + (1 + 2 n) \pi i$ for arbitrary integer $n$. Thus $$ (-2)^{\sqrt{2}} = \exp(\sqrt{2} (\log(2) + (1+2n) \pi i)) = 2^{\sqrt{2}} \exp((1+2n) \sqrt{2} \pi i)$$ In order for this to be real, we would need $ (1+2 n) \sqrt{2}$ to be an integer, and that is not the case because $\sqrt{2}$ is irrational.
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 $2^{\sqrt{2}}(\cos\pi\sqrt{2} + i\sin\pi\sqrt{2})$.