1. Find the sets explicitly

Given $A=\{x\in \mathbb R: |x|<2\}$ so $A=(-2,2).$

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

Solve: $|x-2|\ge 3 \iff x-2\ge 3 \text{ or } x-2\le -3$ $\iff x\ge 5 \text{ or } x\le -1.$ Hence $B=(-\infty,-1]\cup [5,\infty).$

2. Check each option

Option A: $A-B=[-1,2)$

We need elements in $A$ but not in $B$.

Since $A=(-2,2),\qquad B=(-\infty,-1]\cup [5,\infty),$ within $A$, the part belonging to $B$ is only $(-2,2)\cap (-\infty,-1]=(-2,-1].$ Therefore $A-B=(-2,2)\setminus (-2,-1]=( -1,2).$

But option A says $[-1,2),$ which includes $-1$. Since $-1\in B$, it cannot be in $A-B$.

So A is false.

Option B: $A\cup B=\mathbb R-(2,5)$

Compute: $A\cup B = (-2,2)\cup (-\infty,-1]\cup [5,\infty).$ Now $(-\infty,-1]\cup (-2,2)=(-\infty,2).$ So $A\cup B = (-\infty,2)\cup [5,\infty)=\mathbb R-[2,5).$

But option B says $\mathbb R-(2,5)=(-\infty,2]\cup [5,\infty),$ which contains $2$. Since $2\notin A$ and $2\notin B$, it is not in $A\cup B$.

So B is false.

Option C: $A\cap B=(-2,-1)$

Compute intersection: $A\cap B = (-2,2)\cap \left[(-\infty,-1]\cup [5,\infty)\right].$ The second part $[5,\infty)$ has no overlap with $(-2,2)$, so $A\cap B = (-2,2)\cap (-\infty,-1]=(-2,-1].$

Option C gives $(-2,-1)$, which excludes $-1$. But $-1\in A$ and also $|-1-2|=3$, so $-1\in B$. Hence $-1\in A\cap B$.

So C is false.

Option D: $B-A=\mathbb R-(-2,5)$

Compute $B-A$, i.e. elements of $B$ not in $A$.

We have $B=(-\infty,-1]\cup [5,\infty), \qquad A=(-2,2).$ Removing $A$ from $B$:

• From $(-\infty,-1]$, remove $(-2,2)$, leaving $(-\infty,-2]$.
• The set $[5,\infty)$ is unaffected.

Thus $B-A=(-\infty,-2]\cup [5,\infty).$ Now $\mathbb R-(-2,5)=(-\infty,-2]\cup [5,\infty).$ These are equal.

So D is true.

3. Final answer

The only correct option is $\boxed{D}.$