Your question has two main facets. The first is that you did not grasp the way logic does not fall to the liar paradoxes. The second is that there are deeper reasons as to why we have such apparently innocuous sentences in natural language that seem to defy assimilation into formal logic systems.
$
\def\nn{\mathbb{N}}
\def\prov{\square}
\def\t#1{\text{#1}}
\def\pa{\t{PA}}
\def\eq{\leftrightarrow}
$
Why the original liar paradox fails
In classical logic one is only allowed to refer to objects that exist. Because of that, it is impossible to construct the liar sentence, because it is equivalent to:
??? Let $P$ be a sentence such that $P$ is equivalent to $\neg P$.
Which is clearly invalid because we have not proven that such a sentence exists. Indeed, we can show that no such sentence exists. Similarly the barber paradox fails for the same reason, namely that no such barber with the specified property exists.
Why the modified liar paradox fails
For the modified liar paradox, you have made the common error in assuming that provability is a well-defined absolute notion. It is not. Provable in what formal system? So even if we ignore the problem with "this", the string "This sentence is unprovable." is in fact meaningless. It can easily be that "S is unprovable in PA" is true while "S is unprovable in ACA" is false! (Here PA and ACA are two actual formal systems but it is irrelevant here.)
It turns out that given any formal system $S$ with some nice properties, and any sentence $P$ over $S$, there is in fact a sentence denoted by "$\prov_S P$" in the language of arithmetic such that $S$ proves $P$ iff $\nn \vDash \prov_S P$. In English terms, the provability of a sentence over a sufficiently nice formal system is equivalent to whether the collection of natural numbers satisfies some arithmetical sentence. (It is important to note that we are working in some meta-system that has a notion of the collection of natural numbers and understands string manipulation, so that this all makes sense.)
Now this means that a sentence $P$ is unprovable over $S$ iff $\nn \vdash \neg \prov_S P$. Let us see what happens when we try to set up the modified liar paradox in $S$:
Let $G$ be a sentence over $S$ such that $S \vdash G \eq \neg \prov_S G$.
Guess what? Such a $G$ actually exists, which is now called the Godel sentence for $S$. The proof of this fact is the crucial core of Godel's incompleteness theorem.
Yet the paradox vanishes! Let us see what we might try to do within $S$.
Within $S$:
$G \eq \neg \prov_S G$.
If $G$:
$\neg \prov_S G$.
$\color{red}{\prov_S G}$. (WRONG!!!) [Even if $G$ is true, $S$ may be unable to prove "$\prov_S G$".]
[So no contradiction.]
If $\neg G$:
$\prov_S G$.
$\color{red}G$. (ALSO WRONG!!!) [Even if $\prov_S G$ is true, $G$ may not be.]
[Again no contradiction.]
In fact, strangely but truly, given any sentence $φ$ over $S$, if $S \vdash \prov_S φ \to φ$, then $S \vdash φ$. This fact is known as Lob's theorem.
However, $S$ (if sufficiently nice) satisfies the Hilbert Bernay's provability conditions, so the following argument is valid:
$S \vdash G \eq \neg \prov_S G$.
If $S \vdash G$:
$S \vdash \prov_S G$. [by (D1)]
$S \vdash \neg G$.
$S$ is inconsistent.
If $S \vdash \neg G$:
$S \vdash \prov_S G$.
If every arithmetical sentence that $S$ proves is satisfied by $\nn$:
$\nn \vDash \prov_S G$.
$S \vdash G$.
$S$ is inconsistent.
Technical details aside, this is essentially Godel's proof of his first incompleteness theorem, and shows that if $S$ is consistent and is arithmetically sound (namely that every arithmetical sentence proven by it is satisfied by $\nn$), then $S$ neither proves nor disproves $G$. Rosser later proved that the incompleteness theorem holds even if we drop the condition of arithmetical soundness completely, and Kleene found a very elegant computation-theory argument for the strengthened theorem that I detailed in this post.
Why Curry's paradox fails
The above is what is meant when some people say that a correct analysis of the liar paradox yields Godel's incompleteness theorem. Similar analysis of Curry's paradox yields Lob's theorem. I leave this as an exercise for the reader.
Why Quine's paradox fails
People always ask about the liar paradox, but it uses self-reference and so some might say that the problem lies with self-reference. That is not true. There is a deeper paradox in natural language that does not rely on self-reference at all, called Quine's paradox. I shall give the clearest version I can construct here.
It seems that we can always interpret "true sentence" to mean "true statement about reality" unless otherwise specified, and then every sentence is clearly either a true sentence or not a true sentence. Now consider the following sentence $Q$:
" preceded by the quotation of itself is not a true sentence." preceded by the quotation of itself is not a true sentence.
Flawed argument: Notice that $Q$ is a grammatically well-formed sentence of the form X is not a Y., so it ought to be true or false. If $Q$ is a true sentence, then by what it claims, $Q$ is not a true sentence. Therefore $Q$ is not a true sentence. But then by what it claims, $Q$ is a true sentence. Thus we get a contradiction.
Where is the error? Think for a while before continuing!
The answer is that classical logic is essentially based on the fact that there is only one reality, and so when we consider totally precise and unambiguous statements about this reality, it is necessarily the case that every such statement is either true or false about this reality, meaning that either it correctly describes reality, or it does not correctly describe reality. We do not care about statements that are ambiguous or not well-defined, just like we do not care about meaningless nonsense like ß\EÂ{8ÄäÉ5¨5;-c1÷ÌOm¶ÑzYè:ÏÁôví2QêIxú·9Ñ5u¤åÉ¡nçßów⧸}tì-Ì«ÞB8r%sHÛæW¯*".vD.
The key is that we will never be able to justify that $Q$ is a statement about reality, and so we cannot apply the law of excluded middle (LEM) to it. In other words, $Q$ is neither a true sentence about reality nor a false sentence about reality, since it is not even a sentence about reality! (And we can actually justify this claim, as I shall do so below!)
Every sentence can be considered as a string of symbols in reality, so we can rightly say that $Q$ is a sentence, and that "$Q$ is a sentence" is a sentence, but once we want to say something about truth, it may not be a statement about reality anymore. Let us see what we can and cannot say regarding $Q$.
First note that $Q$ is the exactly the same string as " preceded by the quotation of itself is not a true sentence." preceded by the quotation of itself.
Thus $Q$ is equivalent to asserting that $Q$ is not a true sentence. [(*)]
If $Q$ is a true sentence:
$Q$. [We can state $Q$ since it is a true sentence.]
$Q$ is not a true sentence. [By (*).]
Contradiction.
If $Q$ is not a true sentence:
$Q$. [By (*).]
$Q$ is a true sentence. [What we can validly state must be a true sentence.]
Contradiction.
But we do not have LEM for "$Q$ is a true sentence", since we did not prove that it is a statement about reality!
Thus all we can say is that "$Q$ is a true sentence" is not a statement about reality, and likewise $Q$ is not a statement about reality.
Why Berry's paradox and the Surprise test paradox fail
In general the above approach can be used to resolve all logic paradoxes, including Berry's paradox and the Surprise test paradox, which I shall leave as more exercises for the reader. They can also be translated to be in terms of provability, which is instructive to analyze (see this question and the answer).
Let's agree that 'if you are in Paris, then you are in France' ($A \implies B$). (We could get picky, and say maybe you're in Paris, Texas; but let's not!).
But then the converse ($B \implies A$) is not automatically true: for example, we can't then deduce from 'if you are in Paris, then you are in France' that 'if you are in France, then you are in Paris'. In that sense you're right that 'it's not logical' to say the converse is always true.
Now, bear in mind that the converse might be true, or it might not. It must be proven either way based on other information. And it turns out that we know enough other things, about European geography in this case, that we can also prove that the converse is not true.
On the other hand, what we can always deduce is called the contrapositive: once we accept the truth of 'if you are in Paris, then you are in France', then we always automatically can say 'if you are not in France, then you are not in Paris' ($\neg B \implies \neg A$). That will always be true (at least, in the world of mathematical language).
Best Answer
I would like to show you this in truth tables, hopefully it resolves some of the confusion. Of course, you should read the other answers as well, Patrick Stevens answer I like personally.
The connective "$\Rightarrow$" is defined according to this truth table:
$$ \begin{array}{c|l|c|} P & \text{Q} & \text{P $\Rightarrow$ Q} \\ \hline T & T & T \\ T & F & F \\ F & T & T \\ F & F & T \end{array} $$ This should be in your book. As you can see, $P \Rightarrow Q$ is false only when $P$ is true and $Q$ is false, otherwise $P \Rightarrow Q$ is true.
For the contrapositive, we can draw the following table:
$$ \begin{array}{c|l|c|c} P & Q & \neg Q & \neg P & \neg Q \Rightarrow \neg P \\ \hline T & T & F & F & T \\ T & F & T & F & F\\ F & T & F & T & T\\ F & F & T & T & T \end{array} $$
As you can see, for any valuation of $P, Q$ (for any assignment of $T,F$ to $P,Q$), whenever $ P \Rightarrow Q$ is true, so is $\neg Q \Rightarrow \neg P $.
So as you can see, if $\neg Q$ is false, and $\neg P$ is true, then $\neg Q \Rightarrow \neg P$ is true, by the third line in the above table. This is exactly in line with what is says in your book.
For the converse (or inverse) , we have:
$$ \begin{array}{c|l|c|c} P & \text{Q} & \text{P $\Rightarrow$ Q} & Q \Rightarrow P \\ \hline T & T & T & T \\ T & F & F & T \\ F & T & T & F \\ F & F & T & T \end{array} $$
As you can see, they don't match for all valuations, in particular, if $P$ is true, and $Q$ is false, then one implication is true and the other isn't.
Finally: if $A$ is the statement "you are in Paris" and $B$ is the statement "you are in France" and you know $A \Rightarrow B$ is true, then if you are not in France, then you are not in Paris ($\neg B \Rightarrow \neg A$).
And of course, if you are not in Paris ($\neg A$ is true) then you could be on Mars for all we know. You are not necessarily in France.
EDIT: I would like to encourage you to look at more mathematical examples, similar to other peoples' examples in their answers, natural language is full of vague statements that are not always as clear cut as mathematical ones.