Finding the derivative of function with domain empty set

The derivative exists, but as the result of a formal process. For $ \sin^{-1}(e^x + 3)$ put $u=e^x+3.$ Chain rule gives us

$$ \frac{d}{dx}\sin^{-1}(e^x + 3) =\frac{d}{dx}\sin^{-1}(u)= \frac{1}{\sqrt{1-u^2}}\frac{du}{dx}=\frac{e^x}{\sqrt{1-(e^x+3)^2}}.$$

The derivative map didn't even notice the domain or range of $\sin^{-1}(x),$ since in this type of computation we're simply applying an algorithm.

Another thing to notice, is that $e^x+3>3,$ as you pointed out, so the derivative isn't defined either, since the denominator will always be the square root of a negative number.

In a way, what you're touching on here is much deeper. Look at $\ln(x)$ for example. We never consider it as a composition of $x$ with $\ln(x),$ since this seems silly. However, $x$ has domain $\Bbb{R},$ and $\ln$ has domain $\Bbb{R}_{>0}.$ The functions that the derivative map $\frac{d}{dx}$ can evaluate aren't necessarily going to make sense. But as a formal process these will exist, because this is a linear map on a collection of functions which are defined in a formal manner.

You can probably make sense of this as an operation on formal power series, which can be manipulated consistently and are often useful even if they don't converge.

For the power series for $\arcsin$:

Finding the power series of $\arcsin x$

Then substitute the power series for $e^x +3$, expand and differentiate term by term. The answer will be pretty ugly. It will be the formal power series for the "function" you get by naively applying the chain rule to compute the "derivative".

I doubt that this is what the person who wrote the question had in mind. I suspect either a typo, or plain thoughtlessness.

Well, it does not exist in reals, but if you replace $x \in \mathbb{R}$ for $z \in \mathbb{C}$, then you have a well defined function, and $sin(z)$ isn't bounded. It is also easy to see that the same derivative rules work for functions from (an open subset of) $\mathbb{C}$ to $\mathbb{C}$.