If $$A = \{x \in R : |x| < 2\}$$ and $$B = \{x \in R : |x - 2| \ge 3\}$$; then:

We have $$A=\{x\in\mathbb R:\;|x|<2\}.$$

The inequality $$|x|<2$$ means $$-2<x<2,$$ so

$$A=(-2,\,2).$$

Next, $$B=\{x\in\mathbb R:\;|x-2|\ge 3\}.$$

For any real number $$t$$ and positive constant $$c,$$ the rule

$$|t|\ge c \;\Longrightarrow\; t\ge c \;\text{or}\; t\le -c$$

applies. Putting $$t=x-2$$ and $$c=3$$, we get

$$x-2\ge 3 \;\text{or}\; x-2\le -3,$$

that is, $$x\ge 5 \;\text{or}\; x\le -1.$$

Hence

$$B=(-\infty,\,-1]\,\cup\,[5,\,\infty).$$

Let us check each option.

Option A: $$A\cap B$$ is the common part of $$(-2,2)$$ and $$(-\infty,-1]\cup[5,\infty).$$ The only overlap is $$(-2,-1],$$ so $$A\cap B=(-2,-1].$$ Option A claims $$(-2,-1)$$ (without the point $$-1$$), therefore Option A is wrong.

Option B: $$B-A$$ consists of elements in $$B$$ that are not in $$A=(-2,2).$$ Removing the piece $$(-2,-1]$$ from $$B$$ leaves

$$B-A=(-\infty,-2]\,\cup\,[5,\infty).$$

The complement of the open interval $$(-2,5)$$ in $$\mathbb R$$ is exactly

$$\mathbb R-(-2,5)=(-\infty,-2]\,\cup\,[5,\infty).$$

Thus $$B-A=\mathbb R-(-2,5),$$ matching Option B. So Option B is correct.

Option C: $$A\cup B=(-\infty,2)\cup[5,\infty),$$ which omits the point $$2.$$ Option C claims $$\mathbb R-(2,5)=(-\infty,2]\,\cup\,[5,\infty),$$ which does include $$2,$$ so Option C is incorrect.

Option D: $$A-B=(-2,2)-\bigl((-\infty,-1]\cup[5,\infty)\bigr)=(-1,2).$$ Option D states $$[-1,2),$$ which wrongly contains the point $$-1,$$ hence Option D is also incorrect.

Hence, the correct answer is Option B.