Can an event be independent to the Sample Space. Can a Null Event be independent to any other event? How do we show mathematically that event $A$ cannot be independent of $S$ the sample space?
How do we show event $A,$ if $P(A)=0$ cannot be independent of event $B$ if $0 < P(B) < 1?$
Seems like a dumb a question to me. A thing that exists cannot be independent of the universe that it is in, and a thing that does not exist is defined by the universe that it is not. Having tough time with the formulas.
 A: 
How do we show mathematically that event $A$ cannot be independent of $S$ the sample space?

I think it's pretty unlikely that you would be able to show this, given that it's the opposite of true:
Every event $A$ is independent of the sample space $S$
Proof: For any event $A$, $P(A) = P(A \cap S)$ by definition of $S$, and since $S$ is the sample space, $P(S) = 1$. Therefore $P(A \cap S) = P(A) = P(A) * 1 = P(A) * P(S)$, which is the definition of what it means for $A$ and $S$ to be independent events.
Informally, since Event $S$ = "Something happens" describes the whole sample space, we can say "Something always happens, independently of whether any other event $A$ happens."

How do we show event $A$, if $P(A) = 0$ cannot be independent of event $B$ if $0<P(B)<1$?

Again, this is the opposite of true:
Null events are independent of arbitrary events: If $P(A) = 0$ and $B$ is any event, $A$ and $B$ are independent.
Proof: Since $P(A) = 0$, we also have $P(A \cap B) \leq P(A) = 0$, so $P(A \cap B) = 0 = P(A) = P(A) * P(B)$, regardless of what $P(B)$ was. This is the definition of events $A$ and $B$ being independent.
As Teresa Lisbon helpfully reminds us, $P(A) = 0$ means that Event $A$ "practically never happens", and so we can informally restate this as, "If Event $A$ practically never happens, then it practically never happens independently of whether any other event $B$ happens."
A: Taking the definition of independence of events $A$ and $B$ (in the same space)
to be $P(AB) = P(A)P(B),$ you have the following:
If $P(A)=0,$ then $P(AB)=0$ (because $AB \subset A$).
So $P(AB) = P(A)P(B) = 0$ and $A$ and $B$ are independent.
If $A = S,$ then $P(A) = 1$ and $P(AB) = P(B).$ So 
$P(AB) = P(A)P(B) = P(B)$ and $A$ and $B$ are independent.
(See comment by @астон вілла олоф мэллбэрг.)

Sometimes one takes the definition of conditional probability
$P(A|B) = P(AB)/P(B)$ to be the general multiplication rule $P(AB) = P(B)P(A|B),$ but that holds
only for events $B$ with $P(B) > 0$ because division by $0$ is impossible.
Provided that $P(B) > 0,$ it is fine to say intuitively "A independent of B" implies
$P(A|B) = P(AB)/P(B) = P(A)P(B)/P(B) = P(A).$ 

The definition of independence of $A$ and $B$ does "go both ways" in
applications--in the following sense:
(a) If we are making a probability model and we "know" that
two coins are independent, then "getting heads on the first coin
and getting heads on the second" must satisfy $P(H_1H_2) = P(H_1)P(H_2)$
and then if we also "know" the coins are fair, a model consistent with what
we "know" must say $P(\text{Heads on both}) = P(H_1H_2) = (1/2)(1/2) = 1/4.$ 
(b) If we are given a probability model and told
that $P(A) = 1/3,$ $P(B) = 1/2,$ and $P(AB) = 1/6,$ then
$A$ and $B$ must be independent whether or not that seems intuitively obvious.
[Maybe $A$ has to do with occupation and $B$ with income, and we find
it difficult to believe the events can be independent. We can say we
don't believe the given model, nevertheless the given model does imply independence of $A$ and $B$.]
