How many four digit numbers, in which all digits are distinct, contain at least one of 2 and 4 and have no leading zeros? First off, I'm not quite sure what this question is asking (there are a lot of very vague questions in the textbook I'm using and it's caused me some frustration), and I'm assuming one interpretation of the wording.
This is the question:
How many integers between 1,000 and 10,000 are there with distinct digits and at least one of 2 and 4 must appear?
I'm taking it to mean at least one 2 AND at least one 4 must appear.  It can also be interpreted as at least one of either 2 or 4, such that a number containing one two and no fours is valid.  If there is a specific reason why one interpretation is better than the other, please let me know.
Assuming the first interpretation:
Since 10,000 does not include 2 nor 4, it can be ignored and the question boils down to what I put as the title.  To figure out the rest of the question, I scribbled this down (not anything formal):
Let A be a digit whose value is either 2 or 4.
Let B be a digit whose value is neither 2 nor 4.

The different combinations of four digit numbers with distinct digits and at
least one 2 and at least one 4 are represented by the sequences:

BBAA, BABA, BAAB, ABBA, ABAB, AABB

This gives me: $$(7 * 7 * 2 * 1) + (7 * 2 * 7 * 1) + (7 * 2 * 1 * 7) +$$
$$(2 * 8 * 7 * 1) + (2 * 8 * 1 * 7) + (2 * 1 * 8 * 7) = 630$$
The second interpretation doesn't really change the method much, it just results in this instead (10,000 is also ignored here):
Let A be a digit whose value is either 2 or 4.
Let B be a digit whose value is not that of A.

BBBA, BBAB, BABB, ABBB

Which gives me:
$$(8 * 8 * 7 * 2) + (8 * 8 * 2 * 7) + (8 * 2 * 8 * 7) + (2 * 9 * 8 * 7) = 3696$$
Are these answers, and by extension my method of obtaining them, correct?
 A: There are $8 \cdot 7 = 56$ ways of choosing the other two digits. The way of arranging four distinct digits is $4! = 24$. So the answer is $56 \cdot 24 = 1344$.
EDIT To clarify your confusion, the question is asking: how many ways are there are arranging the digits 0,1,...,9 given the restriction that no two digits can appear more than once; and that 2 and 4 must appear (precisely once).
EDIT If we don't allow leading zeros the question is a bit more complicated.
First, calculuate the number when we exclude zero altogether: $7 \cdot 6 \cdot 24$. 
Next, suppose we have a 0,2 and 4 and have freedom only for the final choice. This gives 7 choices for the final digit. There are 24 possible arrangements, but 6 have a zero in the leading position. So there are 18 allowed arrangements. Giving a further $7 \cdot 18$ possible arrangements.
Now $7 \cdot 6 \cdot 24 + 7 \cdot 18 = 1134.$
A: Let us start like this. Suppose 2 and 4 are the first 2 digits. Then we have the number 2 4 _ _. Now, we can do 8*7 to find the other 2 numbers. But we can reaarange these number 4!=24 ways. So the answer is 24*56=1344.
Edit: misread sorry
Edit 2: Sorry, I realized my mistake after reading your solution, I did not mean to try to copy your solution in any way.
Edit 3: since 0 can't be the first digit, doesn't this factor out some cases?
Edit 4: Leading zeroes cannot be permitted, there is no ambuiguity in that matter.
A: My interpretation of the text in the body of the post is that the only bad numbers are those that miss both $2$  and $4$. We calculate the number of good numbers, so that you can compare with your second calculation.
First forget about the $2$, $4$ stuff. If we have no restriction, then the number of numbers with distinct digits, from $1000$ to $9999$, is $(9)(9)(8)(7)$. 
The number of bad numbers with distinct digits (so missing both $2$ and $4$) is $(7)(7)(6)(5)$. 
Subtract. 
