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 $90 + 15 = 105$.
Your ans of $270$ is correct, but you might find this way simpler:
Since a number can't start with zero, there will be $5\cdot6^3$ valid numbers,
of which exactly $1/4$ will be divisible by $4$
[Think of a number system to base 6 to understand the rationale]
Best Answer
I'll answer only some of those questions since they look a lot like homework and no personal work is shown.
1) you can choose the first digit in $5$ ways ($6$ ways if a leading $0$ is allowed), the second digit in $5$ ways (the $4$ digits not used for the first digit and $0$) and the last digit in $4$ ways, so we have a total of $5\times5\times4$ ways without leading $0$ and $6\times5\times4$ with leading $0$
2) if a leading $0$ is allowed you can pick every digit in $6$ ways for a total of $6^3$ ways, if a leading $0$ is not allowed then there are $5\times6^2$ ways (can you see why?)
6) A number is divisible by 5 when its last digit is either $5$ or $0$, so we can pick the last digit in $2$ ways, the second digit in $6$ ways and the first in $5$ or $6$ ways depending on whether leading zeros are allowed, for a total of $2\times6^2$ ways with leading zeros and $2\times5\times6$ without
numbers $4$,$5$ and $7$ can be solved exactly like number $6$ by limiting the last digit to few possibilities, number $3$ is slightly different but you can solve it if you think about the constraints that need to be put on the digits