Having hard time understanding implies $P \Rightarrow Q$ 
I am having hard time understanding the second and third rows in the truth table.
Implies means use if than, but the third statement is confusing. 


*

*$P$ : Tesla Model S is a fast car.  

*$Q$ : Tesla Model S is an expensive car.


*T  $\Rightarrow$ T  : T ( if Tesla is a fast car than it is an expensive car
) ok 

*T $\Rightarrow$ F  : F ( if Tesla is a fast car than it is not an
expensive car )  ok 

*F $\Rightarrow$ T : T ( if Tesla is not a fast car than
it is an expensive car )??? not ok , (why?) (a fast car only should be expensive ???

*F $\Rightarrow$ F      : T ( if Tesla is not
a fast car than it is not an expensive car ) ok



 A: "A implies B" simply means that if A is true, then B must be true as well. Anything that doesn't contradict those criteria is acceptable.
Here's how I think about it: Let's take the statement, "If it is raining, then the driveway is wet"; that is, "it is raining $\implies$ the driveway is wet".
Is it okay for it to be both raining and the driveway to be wet? Yep, that fits. What about neither? Still consistent. Not raining, but wet driveway? That's okay - we never said it was true the other way around. Someone could just be cleaning the driveway off. Raining, but dry driveway? (A dryveway, if you will.) That violates the statement, so it's false. And that's the only way to make the statement false. Everything else is completely consistent with it. And $\implies$ works the same way.
A: Your example statements are not valid; statements from "the real world" are often problematic when dealing with mathematical logic. Your statements are no mathematical statements as "fast" and "expensive" are not well-defined; you might find a car for 500 000 $ expensive whereas I could buy such a car every day and not run out of money. Also I might think that a fast car is any car that can reach 50 km/h, so the meaning of fast/expensive is nothing that can be explained mathematically. 
Therefor I'll use another example that (in my oppinion) is easier to understand the. Also I focus mainly on the third row, as the second row actually shouldn't be a problem (you might want to elaborate on that if you still have problems with this row).

Assuming $x\in\mathbb R$ we have: $$x^2+1=0 \Longrightarrow 2(x^2+1)=0$$

If we have $x\in\mathbb R$ with $x^2+1=0$, then obviously we have $2\underbrace{(x^2+1)}_{=0}=2\cdot 0=0$, so the implication is true. Now let's take a closer look to the statements that are involved:


*

*There exists $x\in\mathbb R$ such that $x^2+1=0$.

*There exists $x\in\mathbb R$ such that $2(x^2+1)=0$.


Our implication can then be written as: $1.\Rightarrow 2.$. But: our first statement is obviously wrong, as $x^2+1=0$ can't be true for any $x\in\mathbb R$. Still as stated above, the implication as a whole is true, regardless of the fact that our premise is wrong.
At first this only looks like an explanation for the fourth row with $1.=F,2.=F$, as by the same logic we can conclude that our second statement is wrong, too. But we never used this fact in either explaining why the implication is true nor in saying that our premise is wrong. We could have used a true statement for $2.$ instead.

Assuming $x\in\mathbb R$ we have: $$x^2+1=0 \Longrightarrow x+x=2x$$

If we have $x\in\mathbb R$ with $x^2+1=0$, then obviously it is true to write $x+x=2x$. Again the implication is true. Let's look at the statements which are involved:


*

*There exists $x\in\mathbb R$ such that $x^2+1=0$.

*There exists $x\in\mathbb R$ such that $x+x=2x$.


Again the first statement is wrong, but the second statement is now true. Now we are in the third row of the truth table, with $1.=F$ and $2.=T$.
I hope this answers your questions, otherwise feel free to ask!
A: By using an implies statement, we try to ascertain the validity of a given conclusion given a hypothesis.
For the proposition if P then Q, P is the hypothesis and Q is the conclusion.
The crucial point is what is the effect of the hypothesis on the conclusion if the hypothesis were to be true?
If the hypothesis is true, its easy to ascertain the proposition's condition (true/false) because it will depend only on the conclusion itself. That is exactly what happens in the first and second statements.
If the hypothesis is false, we can't ascertain the effect of the hypothesis on the conclusion. But the proposition itself doesn't get violated. So the proposition is true.
A: The only case that may be hard to understand is (FALSE => TRUE) = TRUE
Basically, there is nothing to understand in any deep sense:


*

*either you consider that it is a convenient convention and you prove that A => B is equivalent to (NOT A) OR B

*or you consider that => is not a primary connective and you define A => B merely as a shorthand for (NOT A) OR B (check the truth table for this expression)
If you think (FALSE => FALSE) = TRUE also raises a problem, apply the same approach.
As for 2, if TRUE => FALSE was TRUE, you could deduce anything from a single truth.
(Sorry, I don't know how to type the logical signs here.)
