Here is the "slick" way to solve it: there are 3 outcomes for each game (either the odd team wins, they tie, or the even team wins), and there are 3 separate games, so since each game is independent of the other, there are $3^3 = 27$ possible outcomes.
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.
Best Answer
See the wins can be selected in ${8\choose 4}$ ways now losses can be selected in ${4\choose 3}$ ways while only one way is left for tie.so toyal ways are their product=$70.4.1=280$ now another approach consider set{WWWWLLLT} in how many ways can you arrange these letters so its $\frac{8!}{4!.3!}$ which is again equal to $280$ hope its clear .