In order to be less than 4000 you can only use 1, 2, or 3 as the first digit -- so there are 3 ways to choose the first digit.
This leaves you with five digits to choose from for the second digit (you started with six digits to choose from and repetition is not allowed). For the third digit there are four digits to choose from (again, no repetitions), and for the fourth digit there are three digits to choose from.
So overall there are $3 \times 5 \times 4 \times 3 = 180$ numbers.
There are two possible answers to this depending of if the number can start with a 0 or not.
If it can then the number is $7 \times 6 \times 5 = 210$
However we don't usually write numbers with leading zeros. So assuming we don't allow a leading zero the answer is $6 \times 6 \times 5 = 180$
How many of these are odd?
Lets consider picking the numbers in a special order
First we need to pick a 1, 3 or 5 so in our first draw for the final digit so we have a choice of 3. Next we pick our first digit which can be any thing apart from 0 or the number we have just picked making a choice of 5 and for our final draw the middle digit we can pick any of the 5 remaining digits making $3 \times 5 \times 5 = 75$
Finally For the third one greater than 330 we have two ways to achieve this either draw a 4, 5 or 6 for the first digit then we don't care about the others making $3 \times 6 \times 5 = 90$
Alternatively we must draw a 3 for our first digit then 4, 5 or 6 for our second and finally any digit making $1 \times 3 \times 5 = 15$
Now since both of these groups have no overlap we can simply add them together to get $90 + 15 = 105$.
Best Answer
Can't think of nice closed form, but here's a nice linear recursion:
Let $f_n$ be the number of combinations obeying those two rules of length n, and let $g_n$ be the number of those that end with $1$, and $h_n$ be the number that end with $0$. Note that $h_n$ is also the number of such combinations that end with $2$ because of the symmetry, so $f_n$ = $g_n + 2 \cdot h_n$.
But $g_n = f_{n-1}$, because I can always add a $1$ to the end of any such combination of length $n-1$ to get a combination of length $n$, and pulling off a $1$ from the end of such a combination of length $n$ still obeys the two rules. To get a length $n$ combination that ends with $0$, though, the preceding combination has to end with a $0$ or a $1$, so $h_n = g_{n-1} + h_{n-1}$. So, we get $$f_n = f_{n-1} + 2 \cdot \left( g_{n-1} + h_{n-1} \right) \\ = 2 \cdot f_{n-1} + g_{n-1} \\ = 2 \cdot f_{n-1} + f_{n-2}$$