Division by rational (decimal) number meaning When I say, that I exchanged 42 CZK into 1,5 euro. Wh

Short answer:
Multiplication is repeated addition of the same value, so if your are adding some value to itself over and over again, multiplication by the number of times this value occurs in the sum will give the final result. This is where the concept of multiplication comes from.
Division undoes multiplication. This is where it comes from. So if you know what the sum is, and the number of times the value was included in the summation, then the value that was being added to itself can be obtained by division.
(1) If I may be allowed to skip over the added complications of fractional numbers, this is what is happening in your example. Each Euro is worth some number $c$ of CZK. The value of 1.5 Euros is $1.5\times <!--\; \times \; -->c$. But we know that $1.5$ Euros is worth the same as $42CZK$, so we get $1.5\times <!--\; \times \; -->c=42$. We rewrite this as a division:
$c=\frac{42}{1.5}=28$
to see that the value of each Euro is $28$ CPK.
(2) that is the definition of "average": The average of a collection of values is the number that if every one of values in the collection were changed to this number, then the sum of all the values would not change. From this definition, it follows that
$\text{Count}\times <!--\; \times \; -->\text{Average}=\text{Sum}$
and therefore that
$\text{Average}=\frac{\text{Sum}}{\text{Count}}$
(3) "mean" and "average" are the same thing. They are just two different words for the same concept.

Long answer:
Suppose you buy 8 widgets that cost 2 Euro each. How much does it cost? Well, you add them up:
$2+2+2+2+2+2+2+2=16$
That's rather tiring, having to repeatedly add the same thing. So we introduce a second operation: multiplication. Since there are $8$ widgets altogether, we define $8\times <!--\; \times \; -->2=16$. A little development of rules for multiplying, and we have a useful tool that makes a lot of common calculations much easier. But remember that at its base, multiplication is just repeated addition of the same number.
But when you have multiplication, sometimes you end up needing to ask the opposite question: "I bought some wingdings at a price of $1.5$ Euros each, and the total charge was $30$ Euros. How many wingdings did I purchase? (Fortunately, I am buying in a place without sales tax.) If I call the total number of wingdings purchased $N$, then the total cost will be
$\underset{N\text{ times}}{\underset{⏟<!--\; ⏟\; -->}{1.5+1.5+...+1.5}}=N\times <!--\; \times \; -->1.5$
Since I paid $30$ Euros total, it must be that
$N\times <!--\; \times \; -->1.5=30$
So I need some means of undoing that multiplication by $1.5$, so I can figure out what $N$ is. For this reason, we invented division. With additional development of rules and processes, we find convenient means of calculating that
$\frac{30}{1.5}=20$
which means the same thing as saying $20\times <!--\; \times \; -->1.5=30$. So I bought $20$ wingdings. No other number when multiplied by $1.8$ Euros would result a total cost of $30$ Euros.
But in these two examples, everything costs exactly the same amount. Suppose next I buy a combination of 10 widgets, wingdings, and whachamacallits, which cost 2.5 Euros each. The total cost now is 20 Euros. How many did I buy of each? Unfortunately, this question cannot be answered fully. There are several different combinations that would result in 20 Euros:
All widgets: $10\times <!--\; \times \; -->2=20$
8 widgets, 1 wingding, 1 whachamacallit: $8\times <!--\; \times \; -->2+1\times <!--\; \times \; -->1.5+1\times <!--\; \times \; -->2.5=20$
...
5 wingdings, 5 whachamacallits: $5\times <!--\; \times \; -->1.5+5\ast <!--\; \ast \; -->\times <!--\; \times \; -->2.5=20$
Well then, if I can't figure out how many I have of each, is there anything useful I can say? Suppose instead of all costing different amounts, everything had cost the same. What would this amount be, given that $10$ items costs $20$ Euros? We divide to see that if everything cost the same, it would be $2$ Euros.
While this example is contrived, there are a number of cases where there really isn't a good way to control how many items come at one value instead of another. For example, manufacturing variation causes the weight of supposedly identical products to vary from item to item. So you cannot say exactly how much each widget weighs. But, if you weigh 1000 widgets, sum up the weight and divide by 1000 to figure out how much they would weigh if they really were the same weight, then it is likely that the 1000 widgets represents a good sample of the values that might occur from the manufacturing variation. So if you get in another order of 2000 widgets, you can now predict that the weight of this 2000 will be close to twice the weight of the first 1000 even though they don't all weigh the same. Now you don't have to weigh every widget in the future to have a good idea how much any large number of them will weigh.
For this reason, we define the concept of "average" or "mean":
"The average of a collection of values is the number that, if all of the values in the collection were changed to this number, then the sum of the collection would be same."
Now if all of the items in the collection were change to this same average, then the sum of the new changed collection would be the number of items times the average. Since the average is chosen to leave the sum unchanged, we get this formula, we must have that adding the average value to itself the same number of time as there are values in the collection (i.e., multiplying the average by the collection count) gives the original sum of the collection:
$\text{Count}\times <!--\; \times \; -->\text{Average}=\text{Sum}$
Rewriting it using division gives the familiar formula:
$\text{Average}=\frac{\text{Sum}}{\text{Count}}$