I'm unsure how to approach this problem. For a, is it just how many times there are (b, a). So say a = 2, and b = 1, then a forward path would be (1, 2) so there would only be (1, 1), (1, 2) so only 2 forward paths?
I can give you another to look at these. Think of ordinary city-block grid if we move from $(0,0)\to(4,7)$ moving only one block east or one block north at a time. We must move four blocks to the east and seven blocks to the north. Here is a possible walk: $NNENENENENN$ That is a string of seven $N's$ for north & for $E's$ for east.
There are $\dfrac{(7+4)!}{(7!)(4!)}=\dfrac{11!}{(7!)(4!)}$ ways to arrange that string an each arrangement is a possible path.
My answer to your part a) is $\dfrac{(a+b)!}{(a!)(b!)}$
Now you post the rest.