There are two ways of doing this. One is Ross Millikan's: you will make ten "up" moves, and 20 "right" moves; the only question is which order you make them in. Imagine placing the "right" moves on a row; now you need to decide where to do the "up" moves: you do so by inserting them "in between" (or before, or after) the "right" moves. So you need to choose ten places to put "up" moves: there are 21 locations for them (nineteeen in between the "right" moves, one before all of them, one after), and you are allowed to choose the same location more than once.
This is a combinations-with-repetitions: the formula is $\binom{n+r-1}{r}$, where you have $n$ possibilities, and must make $r$ choices with repetitions allowed. In this case, $n=21$, $r=10$, so you get $\binom{30}{10}$.
There is another way of doing it, which is more graphical. I'll do it with a 4 by 3 array so you see how it works. You have this array:
$$\begin{array}{cccc}
\cdot & \cdot & \cdot & \cdot\\
\cdot & \cdot & \cdot & \cdot\\
\cdot & \cdot & \cdot & \cdot
\end{array}$$
Now, you start at the bottom left, so there is only one way to get there; we put a $1$ next to it.
$$\begin{array}{llll}
\cdot & \cdot & \cdot & \cdot\\
\cdot & \cdot & \cdot & \cdot\\
\cdot\;1& \cdot & \cdot & \cdot
\end{array}$$
Then, you can go either up or right; there is only one way to get to those points (via the first move); we put a $1$ next to them:
$$\begin{array}{llll}
\cdot & \cdot & \cdot & \cdot\\
\cdot\;1 & \cdot & \cdot & \cdot\\
\cdot\;1& \cdot\;1 & \cdot & \cdot
\end{array}$$
Now: to get to $(1,1)$, you can either get to it from $(1,0)$ or from $(0,1)$; since there is only one way to get to each of those, there are two ways to get to $(1,1)$. On the other hand, only one way to get to $(2,0)$ or to $(0,2)$:
$$\begin{array}{llll}
\cdot\;1 & \cdot & \cdot & \cdot\\
\cdot\;1 & \cdot\;2 & \cdot & \cdot\\
\cdot\;1& \cdot\;1 & \cdot\;1 & \cdot
\end{array}$$
Next: to get to $(1,2)$, you can arrive either from $(0,2)$ (one way of being there), or from $(1,1)$ (two ways of getting there); so in total, three ways. Likewise, you have three ways to get to $(2,1)$, because you can either go up from $(2,0)$, and there is only one way to do all of that, or you can go right from $(1,1)$ (and there are two ways of doing that, corresponding to the two ways there are to get to $(1,1)$; so we have:
$$\begin{array}{llll}
\cdot \;1 &\cdot\;3 & \cdot & \cdot\\
\cdot\;1 & \cdot\;2 & \cdot\;3 & \cdot\\
\cdot\;1& \cdot\;1 & \cdot\;1 & \cdot
\end{array}$$
Continuing this way, we get:
$$\begin{array}{llll}
\cdot\;1 & \cdot\;3 & \cdot\;6 & \cdot\;10\\
\cdot\;1 & \cdot\;2 & \cdot\;3 & \cdot\;4\\
\cdot\;1 & \cdot\;1 & \cdot\;1 & \cdot\;1
\end{array}$$
So there are $10$ ways to get to the top right corner in the 4 by 3 case.
You may even recognize that these numbers are just Pascal's triangle lying on its side! Well, there is a combinatorial formula for the entries of Pascal's triangle: the $r$th entry in the $m$th row corresponds to the coefficient of $a^{m-r}b^{r-1}$ in the binomial expansion of $(a+b)^{m-1}$, so it equals $\binom{m-1}{r-1}$. To figure out the entry that corresponds to the top right corner, note that you go "down" one row for each position on the $x$-axis, and another one for each step up. So here we have gone to the 4th row on the horizontal steps, and then to the 6th row on the by going up twice. Each step up is a move "right" on the row. So with a 4 by 3, we are in row $4+(3-1)=6$, and we are in position $3$ of that row. According to the formula above, that corresponds to $\binom{4+2-1}{3-1}=\binom{5}{2}$. This corresponds to the need to make $3$ moves right and two moves up, so we need choose where to place the two up moves among the three right moves; there are four places to put them in (before the three, after the three, or in the two spaces in between), so the formula I gave above gives this answer as well.
Look at the $2$d case. If you have a $n_1 \times n_2$ grid and without loss of generality you want to traverse from the top-left corner to the bottom-right, then you must take a total of $n_1 - 1$ steps down and $n_2 - 1$ steps right. The number of sequences of downs and rights is the number of paths and that is where
$$\binom{n_1 - 1 + n_2 - 1}{n_1 - 1}$$
comes from. To generalize this, if you have a $k$-dimensional hypercube with dimensions $(n_1,\ n_2,\ \cdots,\ n_k)$ then you must take $n_i - 1$ steps in each respective dimension, and the number of these sequences of these steps will give you the total paths. The total number is then given by
$$\binom{[n_1 - 1] + \cdots + [n_k - 1]}{n_1 - 1,\ n_2 - 1, \cdots, n_k - 1}$$
which is a multinomial coefficient.
Best Answer
Your reasoning for problem A is right, but you made a careless error, changing $560$ into $960$ at the end. The right answer is 516 triangles. The reason for saying positive area is ust to exclude the sets of three colinear points, as you did.
For problem B, take a broader view: After ten steps, you can end up at any even-total-sum point in the diamond whose corners are $(10,0), (0,10), (-10,0), (0,-10)$. This diamond contains 11 (even) points in the long $y=0$ row, 10 points in each of the $y=1$ and $y=-1$ rows, 9 points in $y=2$ and $y=-2$, and so on down to one point at $y=10$ and one at $y=-10$. The total = $$11 + 2\sum_{k=1}^{10}k = 11 + 2\cdot 55 = 121$$ It is significant that this is $11^2$, and in fact that pattern holds: for $n$ steps there are $(n+1)^2$ possible ending points. This is easily shown by induction.