$X \sim B\left(8, \frac{1}{2}\right) \Rightarrow n=8$ and $p=\frac{1}{2}$ i.e., $q=\frac{1}{2}$
Now $p(|x-4| \leq 2)=p(-2 \leq x-4 \leq 2)=p(2 \leq x \leq 6)$
$\begin{aligned} & \text { i.e., } p(x=2, x=3, x=4, x=5, x=6) \\ & =1-{ }^8 C_0\left(\frac{1}{2}\right)^0 \cdot\left(\frac{1}{2}\right)^8-{ }^8 C_1\left(\frac{1}{2}\right)^1\left(\frac{1}{2}\right)^7-{ }^8 C_7\left(\frac{1}{2}\right)^7\left(\frac{1}{2}\right)^1-{ }^8 C_8\left(\frac{1}{2}\right)^8\left(\frac{1}{2}\right)^0 \\ & =1-\left(\frac{1}{2}\right)^8\{1+8+8+1\} \\ & =1-\frac{18}{256}=1-\frac{9}{128}=\frac{119}{128}\end{aligned}$