Arriving at the conclusion that zero multiplied by undefined is zero I was working on a question which asked whether $f(x).g(x)$ is differentiable at $x=a$, given that $g(x)$ is not differentiable at $x=a$. My solution is as following:

I deduced that $f(x).g(x)$ is not differentiable at $a$.
But then I checked my solution using an example, and it seems that my solution was wrong.

In the above example, $f(x)=x-1$ and $g(x)= (mod(x-1)+1)$. Although $g(x)$ is not differentiable at $x=1$, $f(x).g(x)$ IS differentiable at $x=1$.

And this is how I am arriving at the conclusion that zero multiplied by something undefined is zero (at least in this case).
So my questions are:

Can zero multiplied by something undefined be zero?
If the answer to the above question is no, then what is the mistake in my argument?
Is $f(x).g(x)$ differentiable at $x=a$, when $g(x)$ is not differentiable at $x=a$? What are the necessary conditions for $f(x).g(x)$ to be differentiable at x=a?

Note: $f(x)$ is supposed to be differentiable at x=a.
PS: I have just begun studying calculus, and I am not really a genius in Maths. So please excuse me for any grave msitakes that I may have committed above. Thanks.
Zero multiplied by something undefined is undefined. So the answer is no.
Your mistake lies in a wrong application of the theorem which says: If $h=fg$ and $g$ and $f$ are differentiable at $a$, then $h$ is differentiable at $a$ and $h'(a)=g'(a)f(a)+g(a)f'(a)$. In your case, the hypotheses of the theorem are not fulfilled.
Depends on the situation. You show one example in which none are differentiable or continuous and the product is everywhere differentiable. What is not possible is that $f$ and $g$ are differentiable and $h$ is not. The product rule theorem gives you sufficient conditions, not necessary. Both differentiable, then the product is also. If any one is not differentiable, $h$ could either be or not.
Yes, you showed it yourself! Another fun example would be to look at $$ x\cdot\sin(1/x) $$ at $x=0$.
Your mistake lies in the fact that you have just written down $f(a) g'(a)$. Now, this does not exist but it might be the case that $$ \lim_{x\to a} f(x) g'(x) $$ exists.
Really depends on the situation, but if $f(a)$ is non-zero, it will definitely be not the case. In the other case, you can find examples at both ends.


I hope this helps!