Two chess players, A and B, are going to play 7 games. There are three possible outcomes for each game, A wins, A loses, or Tie

I feel like the chess players should have real names... Archibald and Bartholomew, perhaps? Anyway...

a) A gets $3$ wins, $2$ draws, $2$ losses
This is a question similar to "how many anagrams of "REFEREE" are there?" - the answer is to divide the total arrangements by the arrangements of each repeated element, so

$$N_a = \frac{7!}{3!2!2!} = \frac{5040}{24} = 210$$

We agree :-) - if you calculate out your binomials as factorials, you'll see they cancel down to the same form.

b) A gets exactly $4$ points
Of course we can calculate this by taking cases as above; let's try that:

$$N_b = \frac{7!}{4!0!3!}+\frac{7!}{3!2!2!}+\frac{7!}{2!4!1!}+\frac{7!}{1!6!0!} = 35+210+105+7 = 357$$

c) A gets exactly $4$ points and doesn't lose game $7$
As you can see I've interpreted the condition; basically our final answer can be obtaining by adding the number of ways, after $6$ games, to get to $3$ points, and to get to $3\frac 12$ points - followed by a win and a tie respectively.

Let's just see what that looks like...

$$\begin{align}N_c &= \left( \frac{6!}{3!0!3!}+\frac{6!}{2!2!2!}+\frac{6!}{1!4!1!}+\frac{6!}{0!6!0!} \right )\\ &\quad +\left( \frac{6!}{3!1!2!}+\frac{6!}{2!3!1!}+\frac{6!}{1!5!0!} \right ) \\ &= 20+90+30+1\:+\:60+60+6 = 141+126 \\ &= 267 \end{align}$$

Feel free to check my numbers, as this was done in my head and without the aid of a safety net.