a) contain 3 letters a and the number of the letters b is greater than the number of letters c?
b) all a's are before all b's?
Not sure but I think in the first one it's a permutation with repetition: $$\frac{10!}{3!7!} + \frac{10!}{3!6!1!}+ \frac{10!}{3!5!2!} + \frac{10!}{3!4!3!}$$
For a: You have $3$ a's so you need to pick their places (${10 \choose 3}$ options) and there are 7 spots to fill with two possible letters. There are $2^7$ possibilities, out of which half have more $b$s and half more $c$, so in total: ${10 \choose 3}2^6$.
As pointed out in the comments, your way is almost correct too.
For b: You have $10$ possibilities for the number of $c$s. Suppose you choose to have $k$ of them. Then you need to find places for them (${10 \choose k}$ options) and then decide how to split the remaining $10-k$ letters into $a$ and $b$s. There are $10-k+1$ options ($0,1,...,10-k$) for the number of $a$s and they all must be placed first. In total, the number of words is $$\sum\limits_{k=0}^{10} {10 \choose k}(11-k)=6144$$ where the sum is calculated using known combinatorical indetities.