C(n,k) = n choose k
There are C(5,1) ways to choose the first digit, C(5,1) ways to choose the end digit, obviously no overlap. If even and odd were switched then of course we'd have C(4,1) ways to choose the first digit because 0 would not be an option.
There are 5 digits remaining to be placed, and 8 choices. This makes C(8, 5).
There are 5! ways to permute the central digits.
So I think the overall answer should be C(5,1)*C(5,1)*C(8,5)*5!
The pens are identical, so immediately we can take the last student out of the picture. The equivalent problem states:
How many ways are there to distribute 26 identical pens among three students so that the first student receives at least 3 pens and the second at least two pens?
Maybe you can take it from there.
EDIT: I just noticed something.
"7. How many different 7-digit number begin with an odd digit, end with an even digit, and do not repeat any digit for the remaining three digits?"
Shouldn't that be five instead of three?
EDIT 2: It was a bit silly of me to write C(5,1) above, when I could have just written 5.