To solve for $P\left(X^{2}-12 X+35>0\right)$, we first solve the inequality $X^{2}-12 X+35>0$. This factors as:
$$
(X-5)(X-7)>0
$$
This inequality is satisfied when $X<5$ or $X>7$.
Given the pmf of $X$ :
$$
P(X=x)= \begin{cases}\frac{1}{4} & \text { if } x=2 \\ \frac{1}{2} & \text { if } x=4 \\ \frac{1}{4} & \text { if } x=6\end{cases}
$$
We find:
$$
\begin{gathered}
P(X<5)=P(X=2)+P(X=4)=\frac{1}{4}+\frac{1}{2}=\frac{3}{4} \\
P(X>7)=0
\end{gathered}
$$
Therefore:
$$
P\left(X^{2}-12 X+35>0\right)=P(X<5)+P(X>7)=\frac{3}{4}+0=\frac{3}{4}
$$
