How Implication or Material/Concrete Conditional works when the antecedent is false and the consequent is true When we talk about formal logic, we generally agree that $P$ and $Q$ are deductively valid propositions.
Below fact is supported by Ref. Rutger Uni. page no.2 (PDF doc)
And further when we talk about implication or material/concrete conditional $P\to Q$ we assume: "if $P$ is true, then $Q$ is also true" or more generally in formal logic we assume: if premises are true then conclusions are also true because we are dealing with deductively valid propositions.
But when we draw truth-table of $P\to Q$ we come across one paradox.
That is,

When $P$ is false, and $Q$ is true, $P\to Q$ is still true!

Why so? How can premise (deductively valid proposition or antecedent) be false  with a conclusion (deductively valid proposition or consequent) be true, and their implication be still TRUE!
Reading through the explanations given on the question:
In classical logic, why is $(p\Rightarrow q)$ True if $p$ is False and $q$ is True?
It is bit clear with the given examples that: When $P$ is false, and $Q$ is true, $P\to Q$ is true in some situations.
Out of given examples I am selecting example given by Jai [3rd top answer on that page] which in fact comes from this webpage:

"If you get an A, then I'll give you a dollar."
What if it's false that you get an A? Whether or not I give you a dollar, I haven't broken my promise.

Does this mean that if I get A+ or B- or I was ill for the exam, I will still get a dollar as promised?
Because, I thought propositions are rigid and their use in conditional should be consider as-is without changing the meaning of them, then what is intended by the author of the given example. Here, it is very clear that author of the example suggest that; it is a necessary condition for "Get an A" is "I'll give you a dollar" and a sufficient condition for "I'll give you a dollar" is "Get an A".
Is there any further explanation: The premise (antecedent) requires further support or can be overlooked or ignored in order for conclusion (consequent) to be true when antecedent is false and consequent is true?
 A: You seem to be rather confused ....
You write:

When we talk about formal logic, we generally agree that $P$ and $Q$ are deductively valid arguments (propositions). 

No, we don't agree on that at all.  $P$ and $Q$ are statements, or propositions, or claims.  But they are not arguments!
You also write:

And further when we talk about implication or material/concrete conditional $P\to Q$ we assume: "if $P$ is true, then $Q$ is also true" or more generally in formal logic we assume: if premises are true then conclusions are also true because we are dealing with deductively valid propositions.

In a conditional $P \to Q$, the antecedent is $P$ and the consequent is $Q$. But the antecedent is not a premise, and the consequent is not a conclusion. $P \to Q$ is just a statement, not an argument. And if there is on argument, then there are no premises and conclusion either.
You ask:

But when we draw truth-table of $P\to Q$ we come across one paradox. 
  That is, 

When $P$ is false, and $Q$ is true, $P\to Q$ is still true!

Why so? How can premise (deductively valid proposition or antecedent) be false  with a conclusion (deductively valid proposition or consequent) be true, and their implication be still TRUE!

Again, we are not dealing with any premises or conclusions here, just propositions. and the propositions themselves need not be valid at all.
But, in answer to your question why $P \to Q$ is true when $P$ is false and $Q$ is true: I always like this explanation for why $F \to T$ should be set to True:
Consider the statement $(P \land Q) \to Q$.  Now, that of course should always true since $Q$ logically follows from $P \land Q$.  OK, so now plug in $P=F$ and $Q=T$. Then you get:
$(P \land Q) \to Q = (F \land T) \to T = F \to T$
But like we said, $(P \land Q) \to Q$ should always be true, and so $F \to T=T$
You ask:

"If you get an A, then I'll give you a dollar." 
What if it's false that you get an A? Whether or not I give you a dollar, I haven't broken my promise.
Does this mean that if I get A+ or B- or I was ill for the exam, I will still get a dollar as promised?

No, it does not mean that. If you get an A+ or B-, you may still get a dollar, but you may not.  What is true, however, is that whether you get a dollar or not, the promise has not been broken, and hence "if you get an A, then you get a dollar" is considered true.
In short, the whole conditional is true as soon as its antecedent is false, and in particular you have that if the antecedent is false, and the consequent is true, the conditional is true.
But given the truth of the conditional, if its antecedent is false, that does not mean its consequent is true.
You also ask:

The premise (antecedent) requires further support or can be overlooked or ignored in order to for conclusion (consequent) to be true when antecedent is false and consequent is true?

Again, antecedent $\not =$ premise, and consequent $\not =$ conclusion!
Also, if you know the antecedent is false, in what sense would it be overlooked?  And if the consequent is true, then why would we need to seek further support for it to be true?
I think you are asking one of two questions here (or maybe both):
Why is $P \to Q$ true as soon as $P$ is false, regardless of the value of $Q$?
Why is $P \to Q$ true as soon as $Q$ is true, regardless of the value of $P$?
Well, the previous example can be extended to also cover both the $P=Q=T$ and $P+Q=F$ cases as well. When $P$ and $Q$ are both true:
$(P \land Q) \to Q = (T \land T) \to T = T \to T$
And again, since $(P \land Q) \to Q$ should always be true, we get $T \to T=T$
Likewise, when they are both false:
$(P \land Q) \to Q = (F \land F) \to F = F \to F$
And yet again, since $(P \land Q) \to Q$ should always be true, we get $F \to F=T$
OK, to sum up: we get $F \to T=T$, $T \to T=T$, and $F \to F=T$
The first two should that $P \to Q$ should be true whenever $Q$ is true, no matter the value of $P$.
The first and the third show that $P \to Q$ should be true whenever $P$ is false, no matter the value of $Q$.
I hope that answers all your questions.
A: Classical logic can only be applied to proposition(s) that are unambiguously either true or false at the moment. It has nothing to do with cause and effect, or the passage of time. It deals with things that are true, not things that will be true. I don't think your example fits into that mold.
In any case, for propositions $A$ and $B$, we can prove that $A \implies B$ follows from $\neg A$.
We can prove that $\neg A \implies (A \implies B)$ is a tautology using a truth table. The truth table, however, is based on the standard definition:
$$A\implies B \space\equiv\space \neg (A \land \neg B)$$
Both, however, can be derived using only the following rules of natural deduction:

*

*Conditional proof  ($\implies$ intro)

*Proof by contradiction ($\neg$ intro)

*Joining a pair of statements using conjunction ($\land$ intro)

*Splitting up a conjunction into a pair of statements ($\land$ elim)

*Detachment ($\implies$ elim)

*Removing double negation ($\neg\neg$ elim)


EDIT
See my blog posting on this topic.
