Choosing two numbers from set $\{10,11...,99\}$ that satisfy the given conditions Given the set of numbers $\{10,11,...,99\}$, with no repetitions and no order significance.  
Let $A$ be the set of options of choosing pairs with the same tens number.
Let $B$ be the set of options of choosing only two even numbers. 
Let $C$ be the set of options which the difference between $2$ numbers satisfies $-2 \leq x \leq 2$.
How do I calculate the sizes of $A$, $B$, $C$? 
For $A$: Ss we know we can choose $2$ numbers from $\{10, \ldots, 99\}$,  we have two positions that we need to fill. Therefore, for the first position, we have $90$ possibilities. And after choosing the first number, the second one will only have $9$ options from a group with the same tens number. So that gives us $$\frac{90 \cdot 9}{2}$$
For $B$: In total, we have $45$ even numbers out of $\{10, \ldots, 99\}$. 
For the first position, we have $45$ possibilities. And for the second position, we remain with only $44$ even numbers to choose from.
So that gives us $$\frac{45 \cdot 44}{2}$$
For $C$: For each given number chosen from $\{12, \ldots,97\}$, we can pair it with $4$ different numbers that will fulfill the condition (ex. <12, 10\11\13\14> the subtraction of 12 and all those numbers will give as a difference that is $-2 \leq x \leq 2$.)
And for the numbers $11$ and $98$, there are only $3$ numbers to choose from.
And for the numbers $10$ and $99$, there are only $2$ numbers to choose from. In total: $85 \cdot 4 + 2 \cdot 3 + 2 \cdot 2$.
Is this calculation is right? 
 A: In your attempt, you used the Multiplication Principle.  Since the order of selection does not matter in the first two parts, I will use combinations.
The number of ways of choosing a subset with $k$ elements from a set with $n$ elements is 
$$\binom{n}{k} = \frac{n!}{k!(n - k)!}$$
where $n!$, read "$n$ factorial", is the product of the first $n$ positive integers if $n$ is a positive integer and $0!$ is defined to be $1$. 

In how many ways can two elements be selected from the set $S = \{10, 11, 12, \ldots, 99\}$ that have the same tens digit?

There are nine possible choices for the tens digit.  For each such choice, there are ten numbers with that tens digit, of which we must choose two.  Therefore, the number of ways two elements of $S$ with the same tens digit can be selected is 
$$\binom{9}{1}\binom{10}{2} = \frac{9!}{1!8!} \cdot \frac{10!}{2!8!} = \frac{9 \cdot 8!}{1!8!} \cdot \frac{10 \cdot 9 \cdot 8!}{2 \cdot 1 \cdot 8!} = 9 \cdot 45 = 405$$
Notice that this agrees with your answer since 
$$\frac{90 \cdot 9}{2} = 45 \cdot 9 = 405$$ 

In how many ways can two even numbers be selected from the set $S = \{10, 11, 12, \ldots, 99\}$?

The set $S$ contains $99 - 9 = 90$ elements.  Since the elements of $S$ are consecutive integers, half of them are even.  Hence, set $S$ contains $45$ even numbers.  We can select two of those $45$ even numbers in
$$\binom{45}{2} = \frac{45!}{2!43!} = \frac{45 \cdot 44 \cdot 43}{2 \cdot 1 \cdot 43!} = \frac{45 \cdot 44}{2} = 45 \cdot 22 = 990$$
as you found.

In how many ways can two elements be selected from set $S$ such that the difference of the two numbers satisfies $-2 \leq d \leq 2$?

Given the phrasing of the question, I will assume we are selecting ordered pairs so that $(10, 12)$ has difference $-2$ while $(12, 10)$ has difference $2$. I will also assume that we are choosing two different elements of $S$.
If the first number is $10$, the second number must be one of the two numbers  $11$, or $12$.
If the first number if $11$, the second number must be one of the three numbers $10$, $12$, or $13$.
If the first number is $m$, where $12 \leq m \leq 97$, there are four possibilities for the second number.  They are $m - 2, m - 1, m + 1, m + 2$.  
If the first number is $98$, the second number must be one of the three numbers $96$, $97$, or $99$.
If the first number is $99$, the second number must be one of the two numbers $97$ or $98$.  
Hence, there are $$2 \cdot 2 + 2 \cdot 3 + 86 \cdot 4 = 4 + 6 + 344 = 354$$
ordered pairs of two different numbers in the set $S$ whose difference has absolute value at most $2$.
The only error you made was counting the integers that satisfy the inequalities $12 \leq m \leq 97$.  As indicated in the comments, there are $86$ numbers in the subset $\{12, 13, 14, \ldots, 97\}$ since $97 - 11 = 86$, where we subtract the $11$ positive integers that are not in the subset from the $97$ positive integers that are at most $97$.
