Does it exist a way how to find the ways to get from one point to another when certain points must be avoided in a grid?

I'm assuming the paths must go right and down only.

You can count the paths by using the inclusion-exclusion principle. Without considering the points that must be avoided, there are $\binom{6+6}{6}$ paths. Now subtract the $\binom{4+1}{1}\binom{2+5}{5}$ paths that visit the first bad point, the $\binom{2+4}{4}\binom{4+2}{2}$ paths that visit the second bad point, and the $\binom{4+5}{5}\binom{2+1}{1}$ paths that visit the third bad point. Then add back in the paths that visit two bad points. No paths visit all three bad points.

\begin{align} &\binom{12}{6} -\left(\binom{5}{1}\binom{7}{5} +\binom{6}{4}\binom{6}{2}+\binom{9}{5}\binom{3}{1}\right) +\left(\binom{5}{1}\binom{4}{4}\binom{3}{1}+\binom{6}{4}\binom{3}{1}\binom{3}{1}\right)\\ &=924-(5\cdot 21+15\cdot 15+ 126\cdot 3)+(5\cdot 1\cdot 3+15\cdot 3\cdot 3)\\ &=924-(105+225+378)+(15+105)\\ &=924-708+120\\ &= {\color{red}{366}} \end{align}

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Here's an alternative solution that uses a Pascal-type recursion. Let $p(i,j)$ be the number of good paths that start from $A=(0,0)$ and reach point $(i,j)$. We want to compute $p(6,6)$. By conditioning on the last step into $(i,j)$, we find that $p(i,j)$ satisfies the following recursion: $$ p(i,j)= \begin{cases} 0 &\text{if $(i,j)$ is a bad point}\\ 1 &\text{if $i=0$ or $j=0$}\\ p(i-1,j)+p(i,j-1) &\text{otherwise} \end{cases} $$ The resulting values of $p(i,j)$ are: \begin{matrix} i\backslash j &0 &1 &2 &3 &4 &5 &6 \\ 0 &1 &1 &1 &1 &1 &1 &1 \\ 1 &1 &2 &3 &4 &0 &1 &2 \\ 2 &1 &3 &6 &10 &10 &11 &13 \\ 3 &1 &4 &10 &20 &30 &41 &54 \\ 4 &1 &5 &0 &20 &50 &91 &145 \\ 5 &1 &6 &6 &26 &0 &91 &236 \\ 6 &1 &7 &13 &39 &39 &130 &{\color{red}{366}} \end{matrix} So $p(6,6)=366$.