The translations of "unless" and "except" into symbolic logic. The following two exercises come from Logic for Mathematicians by J.B. Rosser, chapter 2 section one page 17. I am not so sure how to interpret the words "unless" and "except". 
Notation: $\sim P$ represents negation the negation of $P$, and $PQ$ denotes $P\& Q$ which the author refers to as the logical product of $P$ and $Q$. Also, $P\equiv Q$ denotes "$P$ if and only if $Q$".
Exercise 3. If “$P$
 ”, “$Q$
 ”, and “$R$
 ” are translations for “$x=y$
 ” , “$x/z=y/z$
 ”, and "$z\ne0$"
 , write a translation for “if $x=y$
 , then $x/z=y/z$
  except when $z=0$
 ” 
Solution: It is not the case that $(P\rightarrow Q)\sim R$
 , i.e. we have $\sim[(P\rightarrow Q)\sim R]$
  which is equivalent to $\sim[\sim(P\sim Q)\sim R]$
  which is equivalent to $\sim\{[\sim P\sim(\sim Q)]\sim R\}$ (Note: the last step was an invalid manipulation, hence my conclusion is wrong)
 . Since $\sim(\sim Q)$
  is equivalent to $Q$
  and because the logical product is both associative and commutative, we have $\sim[Q\sim(PR)]$
  hence $Q\rightarrow RP$.
My confusion is that I am pretty sure that $Q\equiv RP\space$  holds but i don't think that it would be the right translation. Am I right?
Exercise 4. If “$P$
 ” and “$Q$
 ” are the translations of “$(n-1)!+1$
  is divisible by $n$
 ” and “$n$
  is prime” then write a translation for “$(n-1)!+1$
  is not divisible by $n$
 , unless $n$
  is prime”
Solution: In other words we have that “the only time $(n-1)!+1$
  is divisible by $n$
  when $n$
  is prime” or “$(n-1)!+1$
  is divisible by $n$
  when and only when $n$
  is prime”, hence $P\equiv Q$.
I am not sure if I interpreted `unless' correctly here.

More Reasoning for Exercise 3: 
I arrive at the conclusion  $\sim[(P\rightarrow Q)\sim R]$ because it says "$P\rightarrow Q$ except when $\sim R$", which I take to mean that "it is false that both $\sim R$ and $P\rightarrow Q$ are true at the same time"; i.e. it is not the case that $(P\rightarrow Q)\&\sim R$, hence $\sim[(P\rightarrow Q)\sim R]$. From there mechanical manipulation of symbols gets me to $Q\rightarrow RP$ (again this is wrong).
Additionally, though I am certain that $RP \rightarrow Q$ I am not quite sure how to reason this from the sentence “if $x=y$, then $x/z=y/z$  except when $z=0$”. Furthermore, with my interpretation of it being $\sim[(P\rightarrow Q)\sim R]$ I am not sure what symbolic manipulations would give $RP \rightarrow Q$.
 A: Considering "unless" means "without", "if not", we have several ways to translate "$P$ unless/except $R$". Note that


*

*Without $R$, $P$ is true. We can write "$\sim R\to P$".

*Either $P$ or $R$ is true. We can write "$P$ or $R$"

*If $P$ fails, $R$ is the reason. We can write "$\sim P\to R$"


Moreover, "$P\to Q$" is logically equivalent to "$\sim P$ or $Q$", also "$\sim(P$ and $\sim Q)$"
So don't confuse by what $P,R$ is. They are only symbols representing propositions.

For Ex3, denote if part "$P\to Q$" by $S$ and unless part "$\sim R$" by $T$, then it's routine to write
$$\sim T\to S\equiv\sim(\sim R)\to(P\to Q)\equiv\sim(R\sim(\sim(P\sim Q)))\equiv\sim(R(P\sim Q))$$
Ex4 is similar and more straightforward.

Yes, sometimes we explain "$P$ unless $Q$" as $P\equiv\sim Q$. That means $P$ is true if $Q$ is not and vice versa. But that depends.
If "$P$ unless $Q$" means "$P\equiv\sim Q$", then whenever $Q$ is true, $P$ cannot be true, or "$Q\to\sim P$". Let's check it for Ex3. Assume $z=0$ and $x=y$, we require $x/z=y/z$ is false. However, $x/z=y/z$ is neither true nor false because it's undefined. And Ex4, similarly we assume $n$ is prime, it's expected that $(n−1)!+1$ is not divisible by $n$ is false, or $n\vert (n-1)!+1$ and so is the fact, in which case you can say "$P\equiv\sim Q$".
In above examples, we see "unless" can have different meanings in different context. However for propositional logic, we don't have any context. No one knows what a proposition symbol represents. That's why in propositional logic, "unless" means "if not" instead of "if not and only if not".
A: I will follow the suggestion of S.C.Kleene, Mathematical Logic (1967), where he summarize in a table a list of expressions and their possible translation in symbols (pag.64) :

$A \lor B$ is $A$ unless $B$ [usually] and is $A$ except when $B$ [usually].

Try now with the exercises :
Exercise 3 : If $P$, $Q$ and $R$ are translations for “$x=y$” , “$x/z=y/z$” and "$z = 0$", write a translation for “if $x=y$, then $x/z=y/z$ except when $z=0$” 
I will start from the mathematical condition, rewritten as : $\lnot z = 0 \rightarrow (x=y \rightarrow x/z=y/z)$ i.e. $ z = 0 \lor (x=y \rightarrow x/z=y/z)$.
We have that : $R \lor (P \rightarrow Q)$, i.e. $(P \rightarrow Q) \lor R$.
We can read it as : “if $x=y$, then $x/z=y/z$, except when $z=0$”
Exercise 4 :  If $P$ and $Q$ are the translations of “$(n-1)!+1$ is divisible by $n$” and “$n$ is prime” then write a translation for “$(n-1)!+1$ is not divisible by $n$, unless $n$ is prime”
The mathematical condition is : [$(n-1)!+1$ is divisible by $n$] $\rightarrow$ ($n$ is prime) i.e. $\lnot [(n-1)!+1$ is divisible by $n$] $\lor$ ($n$ is prime).
We have that : $\lnot P \lor Q$.
A: Definitions: 
except: with the exclusion or exception of 
unless: 1) except on the condition that : under any other circumstance than 2) without the accompanying circumstance or condition that
Retrieved from here
Exercise 3:
Solution: The first step to forming our translation is to strip the statement of its meaning, we notice that our statement is of the form “$P\rightarrow Q$
  except when $\sim R$”. Now we are left to interpret what “except when” means. With our definition above we can write “$P\rightarrow Q$
  excluding the case $\sim R$” which can be interpret two ways “When $\sim(\sim R)$
  is true then $(P\rightarrow Q)$
  is true” in symbols this says “$\sim(R\sim(P\rightarrow Q))$”, i.e. $R\rightarrow(P\rightarrow Q)$. We could have also interpret the statement as “It is not the case that $\sim R$
  and $P\rightarrow Q$
  are both true” which is simply to say that “$\sim(\sim R(P\rightarrow Q))$
 ” . i.e.$(P\rightarrow Q)\rightarrow R$. If we were to consider the conjunction of both of the interpretations above we would then have to conclude “$P\rightarrow Q$
  except when $\sim R$
 ” is to be translated as “$(P\rightarrow Q)\equiv R$”. 
Yet another way that we could translated “$P\rightarrow Q$
  except when $\sim R$
 ” is to say “$\sim(P\rightarrow Q)$
  when and only when $\sim R$
 ” or “$P\rightarrow Q$
  when and only when $\sim(\sim R)$
 ”; in either case we have to conclude again “$(P\rightarrow Q)\equiv R$”. Our conclusion is also in line with the following source on slide 114; moreover, I have found another source here on page 157 that also encourages us to be careful with the word “unless”. As explained in the first source we should discern between a strong “unless” and a weak “unless”, the strong “$P$
  unless $Q$
 ” is to be translated as “$P\equiv\sim Q$
 ” while the weak version is to be interpreted as “$\sim Q\rightarrow P$
 ” (as suggested by many in the other answers). The author goes on to explain that when in doubt we should assume the weak “unless” is being used. However, when trying to demonstrate the strong form of “unless” the author chose to substitute “unless” for “except” which might lead one to believe that “except” implicitly mean strong “unless” as I have concluded above.
Nonetheless, this is only speculation on my part from the little information given by those whom I assume to know more about the subject than myself. I can only conclude from this exercise that “except” and “unless” are interjected into our logic because of common English, that is,at times we have no choice but to deal with the wording that we are presented. However, there are many alternatives for these word which eliminate ambiguity, and thus we should strive to avoid the use of the "unless" and "except" for they are sloppy due to their nature.
Exercise 4:
To begin our statement says “$\sim P$
  unless $Q$
 ”, in which case we usually consider a “unless” which is to be interpreted as “If not $Q$
  then $\sim P$
 ” or “$\sim Q\rightarrow\sim P$
 ”, which is equivlent to “$P\rightarrow Q$
 ”. (The preceding was explained in detail in Exercise 3). However from context it is also clear that “$Q\rightarrow P$
 ”, in which case we have “$P\equiv Q$
 ”. Nonetheless as a rule of thumb we will always interpret “$P$
  unless $Q$
 ” as “$\sim Q\rightarrow P$
 ” and not as “$P\equiv\sim Q$
 ” which is equivelent to “$\sim P\equiv Q$”.
Now on a side note, in my original question I claimed that $R\rightarrow(P\rightarrow Q)$  is equivalent to  $RP\rightarrow Q$. I also claimed that   $(P\rightarrow Q)\rightarrow R$ was logically equivalent to  $Q\rightarrow RP$  which I disprove in the truth table below:

Additionally I added the following truth table to confirm $RP\rightarrow Q$:

A: 
Write a translation for "if $x=y$, then $x/z=y/z$  except when $z=0$".

This can also be written, "if $z\neq 0$ then (if $x=y$, then $x/z=y/z)$".
Using the implication operator, we have 
$z\neq 0 \implies [x=y \implies x/z = y/z]$.
But you cannot use implication, only conjuction and negation. By definition, 
$A\implies B \equiv \neg[A \land \neg B]$
Applying this definition twice, we would have 
$\neg[z\neq 0 \land [x=y \land \neg [x/z= y/z]]]$
which can be easily translated into your special notation as 
~(R(P(~Q)))
A similar analysis applies to the other exercise. Note that, as used here, "unless" and "except when" are interchangeable. 
A: Please, note that the Dover edition of J.B.Rosser, Logic for Mathematicians (2008) has (pag.17) in Ex.II.1.3 : $R$ for "$z = 0$".
