Find the probability that among seven persons no two were born on the same day of the week I Was initially thinking P(born on Monday) = 1/7 and P(Not Born on Monday) = 6/7 and then 1/7 * 6/7 = 6/42, but I don't know if that's the correct approach? 
In addition what is P(at least 2 were born on same day) 
and P(two were born on a Saturday and 2 born on Tuesday) ? 
 A: This problem could be a bit more complicated, but the fact that there are 7 people and 7 days of the week makes it simpler. That's because, in this case, each day of the week must be assigned to exactly one person. There are $7^7$ ways to assign days of the week to 7 people. How many ways are there to assign days of the week with no repeats? You can think of looking at a calendar week and trying to slot the people in. It's just like ordering 7 books on a shelf. There are 7 choices for who can be born on Sunday, then 6 left for Monday, and so on. Take that number and divide it by $7^7$, and you're done.
As @Bram28 says in the comments, this allows you to easily calculate the probability that the opposite happens -- that some two were born on the same day. 
I'll leave it to you to figure out the last question.
A: 
I Was initially thinking P(born on Monday) = 1/7 and P(Not Born on Monday) = 6/7 and then 1/7 * 6/7 = 6/42

OK, let's think about this. How is your P(born on Monday) defined? That is for any random person, right? And so sure, P(Not Born on Monday) = 6/7.  But: if you multiple those (which, BTW, would not be $\frac{6}{42}$, but $\frac{6}{49}$), what does that mean?  It is the probability that for some random person, that person is born on a Monday and not on a Monday? But that is assuming these two events are independent .. which clearly they are not.  Is it that for two people, the first is born on a Monday, and the second one is not? Those are independent, sure ... but we have $7$ people involved, not just $2$. 
Here is what you need to do: you need to line up all $7$ people (call them $A$ through $G$), and then do something like this: OK, it doesn't matter what day $A$ is born, but we need to make sure $B$ is born on a different day than $A$ was, the probability of which is $\frac{6}{7}$. OK, now assuming that $B$ is indeed born on a different day than $A$, two weekdays are taken, but we need to make sure $C$ is born on a different day yet, the probability of which is $\frac{5}{7}$. ... see where this is going?
