1) The even position are $n$, so there are $n! \cdot n! $ such permutations, because even numbers must stay in even positions and odd numbers in odd positions.
2) The answer is the total number of permutations minus the number of permutations in which all even numbers are in odd positions, so $$(2n)!- n!\cdot n!$$
How many six-digit positive integers contain exactly two even digits?
Remember that an integer $n$ is said to be even if there exists an integer $m$ such that $n = 2m$. An integer that is not even is said to be odd. Hence, there are five even digits: $$0, 2, 4, 6, 8$$ and five odd digits: $$1, 3, 5, 7, 9$$
Your strategy of breaking the problem into two cases is correct. Let's correct your calculations for those cases.
Case 1: The leading digit is even.
Since the leading cannot be zero, there are four ways to fill the leading digit. There are five ways to choose the position of the other even digit and five ways to fill that position with an even digit. There are five ways to fill each of the four remaining positions with an odd digit.
$$4 \binom{5}{1} \cdot 5 \cdot 5^4 = 4 \binom{5}{1} 5^5$$
Case 2: The leading digit is odd.
Two of the remaining five positions must be filled with even digits. There are $\binom{5}{2}$ ways to select the positions of the even digits and five ways to fill each of those positions with an even digit. There are five ways to fill each of the four other positions with an odd digit.
$$\binom{5}{2} \cdot 5^2 \cdot 5^4 = \binom{5}{2} 5^6$$
Total: Since the above cases are mutually exclusive and exhaustive, the number of six-digit positive integers that contain exactly two even digits is
$$4 \binom{5}{1} 5^5 + \binom{5}{2} 5^6$$
In how many five-digit positive integers are there digits that appear more than once?
Hint: Subtract the number of five-digit positive integers in which all the digits are distinct from the total number of five-digit positive integers.
The total number of five-digit positive integers is $$9 \cdot 10 \cdot 10 \cdot 10 \cdot 10 = 90000$$ since the leading digit can be chosen in nine ways (as $0$ is prohibited) and each of the remaining digits can be selected in $10$ ways. Alternatively, the largest five-digit number is $99999$ and the largest number with fewer than five digits is $9999$, so the number of five-digit positive integers is $$99999 - 9999 = 90000$$ The number of five-digit positive integers with distinct digits is $$9 \cdot 9 \cdot 8 \cdot 7 \cdot 6 = 27216$$ since we have nine choices for the leading digit (as $0$ is prohibited), nine choices for the thousands digit (since we cannot use the leading digit), eight choices for the hundreds digit (since we cannot use the tens thousands or thousands digits), seven choices for the tens digit, and six choices for the units digit. Therefore, the number of five-digit positive integers that do contain a digit that appears more than once is $$9 \cdot 10 \cdot 10 \cdot 10 \cdot 10 - 9 \cdot 9 \cdot 8 \cdot 7 \cdot 6 = 90000 - 27216 = 62784$$
Best Answer
For me, it often helps to think of sample objects which I am counting.
One of your confusions seems to be what $a \times b$ means in a counting formula. I like to think of it as choosing among $a$ options first, and then, choosing among $b$ options, in that order (if order matters). A typical object being chosen is a $2$-vector $(A,B)$, where $A$ is one of the $a$ options and $B$ is one of the $b$ options. In particular a typical object is not a $2$-set $\{A, B\}$.
For ${50 \choose 2} \times {49 \choose 1}$, a typical object is a $2$-vector $(A,B)$ where $A$ is a $2$-subset of odd numbers and $B$ is a $1$-subset of even number. Every desired $3$-subset of $2$ odd numbers and $1$ even number can be written in this way in exactly one way. So no need to divide. E.g. $\{23, 4, 17\} = (\{23, 17\}, \{4\})$.
For $50 \times 49 \times 49$, first decide which $49$ means what. The fact that they are the same number is just a coincidence. Lets say you arrived at that formula thinking the first $49$ is choosing the second odd number, i.e. you're choosing odd-odd-even. So then a typical object is a $3$-vector $(A,B,C)$ where $A, B$ are distinct odd numbers and $C$ is even. Any desired $3$-subset can be written in this $3$-vector form in $2$ ways, e.g. $\{23, 4, 17\} = (23,17,4)$ or $(17,23,4)$, and crucially, $(23,17,4) \neq (17,23,4)$, so you divide by $2!$.