Determining the numbers of subsets in the set of naturals $\{1,2,...,100\}$ that check two conditions 
Let $X$ be the set $\{1,2,...,100\}$. Determine the number of subsets $\{x,y\}$ such that:

*

*At least one between $x$ and $y$ is even

*$3\mid x+y$

Points $1$ and $2$ are two separate cases. I'm not looking for the number of subests that fulfill both conditions, but rather for the number of subsets that fulfill the first condition, then the number of subsets that fulfill the second one.
I have some ideas but I'm not sure my reasoning is correct.
As for the first point, we have that there are $50$ even numbers in $X$. We fix an even number, and then we have $99$ ways left of choosing the second member of the subset. Therefore, the answer is: $$50\cdot99.$$
As an aside: if this would've been about ordered pair, instead of subset, I think I should have doubled that and then subtract the pairs where both numbers are even, because they got counted twice. So that would have been: $$2\cdot50\cdot99-{50\choose2}.$$
Are these two solutions adequate?
For the second point, I noticed that, fixed a number, other numbers that, when added, equal a multiple of $3$ are spaced $2$ numbers apart. Also, the first number I can choose to get a multiple of $3$ depends on the first one I fixed.
For numbers $1,2,3$, these are respectively $2,1,3$, then they repeat periodically.
I'll have $100$ ways of choosing the first number, of course. The second one will have initially about one third of the cardinality of $X$ ways to be choose, since one in three numbers is good.
However, I'm not sure how to exactly quantify this. What am I missing?
 A: A very good way of testing if an argument holds is to apply it on a case that's small and can be verified by hand. So for the first question, let's go back from the set $\{1,2,\ldots,100\}$ to the managable $\{1,2,3,4\}$.
By your argument, you now have $2$ options to choose the even number, then $3$ remaining choices for the second member of the set, resulting in $2\times3=6$ sets that have at least one even element.
If you look at the 2-element subsets of  $\{1,2,3,4\}$, you easily enumerate them, there 6:
$$\{1,2\}, \color{red}{\{1,3\}}, \{1,4\}, \{2,3\}, \{2,4\}, \{3,4\},$$
and $5$ of them fulfill your condition (all except the red one). So you predicted $6$ such sets, but only $5$ exist. So something in your argument is incorrect, but what?
Again, because the numbers involved are small, you can actually write down how you arrived at the prediction! You argued you can can choose one of 2 even numbers first ($2$ or $4$), then have 3 remaining numbers for the second element. That boils down to
$$\{2,1\}, \{2,3\}, \color{blue}{\{2,4\}}$$
if you choose $2$ first and
$$\{4,1\}, \color{blue}{\{4,2\}}, \{4,3\}$$
if you choose $4$ first. 
Now note that you have listed the set $\{2,4\}=\{4,2\}$ twice, which explains why you overcounted by 1. Now, seeing that both number are even, maybe you can find the flaw in your argument and correct it...

For your second question, consider the 3 sets
$$R_0=\{3,6,9,12,\ldots,96,99\}, R_1=\{1,4,7,10,\ldots,97,100\}, R_2=\{2,5,8,11,\ldots,95,98\}.$$
The first set contains all numbers from $\{1,2,\ldots,100\}$ that are divisible by $3$, the second all numbers from $\{1,2,\ldots,100\}$ that leave remainder $1$ when divided by $3$ and the third set all numbers from $\{1,2,\ldots,100\}$ that leave remainder $2$ when divided by $3$. 
First convince yourself that to know if $3|(x+y)$ you don't need to know what $x$ and $y$ are exactly. It is enough to know from which of the above sets $x$ comes and from which $y$ comes. For example, if $x\in R_1$ and $y\in R_2$, then $(x+y)$ will indeed be divisble by $3$. OTOH, if $x\in R_0$ and $y\in R_2$, $(x+y)$ will leave remainder $2$ when divided by $3$, so $3\nmid(x+y)$.
Now, considering that you can easily count how many elements $R_0, R_1$ and $R_2$ have, can you find out how many subsets there are fullfulling your condition 2?
