Yeah, there is a nice way to do it. This looks long, but it's because I stated everything rigorously. If you draw pictures while reading this, it'll make a LOT more sense.
Let $f(2n)$ denote the number of paths from $(0, 0)$ to $(2n, 2n)$ not crossing through a point of the form $(2k+1, 2k+1)$. I claim that $f(2n) = C_{2n}$, where $C_{2n}$ is the $2n$-th Catalan number.
A well known property of Catalan number $C_{n}$ is that it satisfies the following recursion formula:
$$ C_{n+1} = \sum_{i=0}^{n} C_i C_{n-i} \tag{1}$$
Another well known property is that it counts the number of paths from $(0,0)$ to $(2n,2n)$ which never go above the line $y=x$.
I'll prove the result by induction. Notice it's true for a base case of $n = 0$. Now, suppose the result is true for $f(0), f(2), \dots, f(2n-2)$.
To count $f(2n)$, we do casework on the first point of the form $(2k, 2k)$ our path goes through (other than $(0, 0)$). This casework covers all paths since all paths end up at $(2n, 2n)$. Suppose the first such point is $(2k, 2k)$. WLOG on our first step, we went $(0, 0) \to (1, 0)$, we'll multiply by $2$ in our final count. Then we must also end with $(2k, 2k-1) \to (2k, 2k)$. It remains to count the number of paths which go from $(1, 0)$ to $(2k, 2k-1)$ without passing any point of the form $(2k, 2k)$. This is just $C_{2k-1}$! After this, there are $f(2n-2k)$ ways to finish up the path $(2k, 2k) \to (2n, 2n)$. Therefore, we have
$$f(2n) = \sum_{k=1}^{n} 2 \cdot C_{2k-1} f(2n-2k)$$
By the inductive hypothesis, $f(2n-2k) = C_{2n-2k}$, so we really have
$$f(2n) = \sum_{k=1}^{n} 2 \cdot C_{2k-1} C_{2n-2k} = \sum_{k=1}^n C_{2k-1}C_{2n-2k} + \sum_{k=1}^nC_{2k-1}C_{2n-2k}$$
using $j = n-k$ as the iterator for the second sum, we get
$$f(2n) = \sum_{k=1}^n C_{2k-1}C_{2n-2k} + \sum_{j = 0}^{n-1} C_{2j} C_{2n-2j}$$
The finish is in sight! The first sum is just $C_1C_{2n-2}+C_3C_{2n-4} + \dots C_{2n-1}C_{0}$ (i.e. the odd terms from $(1)$) while the second sum is just $C_{0}C_{2n-1} + \dots C_{2n-2}C_1$ (i.e. the even terms from $(1)$). Therefore, we deduce that $f(2n) = C_{2n}$ as desired.
I'm sure the bijective proof exists, but I have yet to try to find it. But given this, maybe you'll be able to do it :)
You need to look at the complement, which is all the paths from $(0,0)$ to $(7,5)$ that pass through $(3,3)$.
As you noted, the number of paths from $(0,0)$ to $(3,3)$ is ${6\choose3} = 20$, and the number of paths from $(3,3)$ to $(7,5)$ is ${6\choose4} = 15$. So the number of paths from $(0,0)$ to $(7,5)$ that pass through $(3,3)$ is all combinations of the above paths, which means $20\cdot15=300$ paths.
$792-300=492$, and so $492$ is the final answer.
Best Answer
An alternative to inclusion-exclusion is to use recursion. Let $p(x,y)$ be the number of such paths from $(0,0)$ to $(x,y)$. By considering the last step into $(x,y)$, we find that $$p(x,y)=p(x-1,y)+p(x,y-1),$$ where $p(x,y)=0$ if $x<0$, $y<0$, or $(x,y)$ is blocked. The boundary condition is $p(0,0)=1$, and you want to compute $p(10,10)$.
\begin{matrix} 1 &11 &66 &166 &441 &1283 &3608 &7416 &14410 &29078 &\color{red}{60256} \\ 1 &10 &55 &100 &275 &842 &2325 &3808 &6994 &14668 &31178 \\ 1 &9 &45 &45 &175 &567 &1483 &1483 &3186 &7674 &16510 \\ 1 &8 &36 &0 &130 &392 &916 &0 &1703 &4488 &8836 \\ 1 &7 &28 &64 &130 &262 &524 &980 &1703 &2785 &4348 \\ 1 &6 &21 &36 &66 &132 &262 &456 &723 &1082 &1563 \\ 1 &5 &15 &15 &30 &66 &130 &194 &267 &359 &481 \\ 1 &4 &10 &0 &15 &36 &64 &64 &73 &92 &122 \\ 1 &3 &6 &10 &15 &21 &28 &0 &9 &19 &30 \\ 1 &2 &3 &4 &5 &6 &7 &8 &9 &10 &11 \\ 1 &1 &1 &1 &1 &1 &1 &1 &1 &1 &1 \\ \end{matrix}