Can we talk about the continuity/discontinuity of a function at a point which is not in its domain? Let us say that I have a function $ f(x)=\tan(x)$ we say that this function is continuous in its domain.
If I have a simple function like $$ f(x)=\frac{1}{(x-1)(x-2)} $$
Can we really talk about its continuity/discontinuity at $x=1$ or at $x=2$.
From what I know we can't since it is not in its domain.
But doesn't it make every function of the form $$ f(x)=\frac{1}{g(x)}$$ continuous in its domain. Where $g(x)$ is any polynomial and $g(x)=0$ at n points (lets say).

A function by definition must be defined for all points in the domain. So formally speaking a function like $\tan(x)$ doesn't even know what $\pi/2$ is (other than the codomain maybe). So no, it does not make sense to talk about continuity at points outside the domain. 
For example is the function $f\colon \Bbb{R} \to \Bbb{R}$, $x\mapsto x$ continuous at the point $x=\text{New York}$? It is just as meaningless to talk about $\tan(x)$ being continuous at $\pi/2$ as it is to talk about it being continuous at $\text{New York}$.
Another way to see this is the definition. A function is continuous at a point $a$ iff:
$$\lim\limits_{x\to a} f(x) = f(a)$$
Well if the right side doesn't exist, this clearly can't be true.
