Question: Are there any set of numbers into which any of the indeterminate forms we see in a calculus course, like 00, n/0, 1infinity, etc has an answer?

I'm asking that because, thanks to the Net, I took notice of other kinds of numbers besides those commonly seen in the high school and most of the university courses: Real and Complex numbers.

There are Hyperreals, Surreals, Quaternions, ... so I thought that some indeterminate form could be have an answer among them, just like we have an answer for the x-th root, x being par, of negative numbers in the set of complex numbers.

Answer: 
Indeterminate forms are *indeterminate* because they could be defined in a different manner and reach different values.

The simplest example is $0\cdot\infty$. This is indeterminate because it depends on what $0$ means and what $\infty$ means.

Notice that in all indeterminate form we take that values may be defined through different limits, $0$ does not have to be $0$ but some expression that evaluates to $0$. So in essence they are indeterminate because some common operations like switching limits does not yield the same result - necessarily, or similar other limitations.

If you however take $0$ literary as $0$ then anything times $0$ is $0$. So from this perspective there is nothing that is undetermined. However, in this case we have defined that $0$ means $0$. This is actually what is meant when we say an indeterminate form. It says: the expression depends on the definition of $0$ and $\infty$.

So no, it is not the situation like $\sqrt{-1}$ where this expression is not a part of the set of real numbers.

It is like the situation when you specify an additional condition for the expressions like $\frac{\sin(x)}{x}$ for $x=0$ in order to achieve a goal of having a nice behaving function.

Take for example that $0 \cdot \infty=5$. This requires $0=\frac{5}{\infty}$ Take a function that reaches $0$ for whatever like $f(1)=0$ and have $\frac{5+f(x)}{f(x)+\infty}$ I ask you: what is $\lim\limits\_{x \to 1} \frac{5+f(x)}{f(x)+\infty}$?

There is the catch.

This last expression evaluation totally depends on the game between $f(x)$ and $\infty$. It can be $0$ but it does not have to be.
$0^0$ is an interesting example.

Since

$$\lim\_{x->0} x^x = 1$$

it makes sense to define 

$0^0 := 1$

This also has advantages concerning the binomial theorem.

But some mathematicians set $0^0 := 0$ and some say it is
indetermined. There seems to be no consense what $0^0$ should be.

The other examples you mentioned are even more problematic.