Let $$A =\left\{x: |x^{2}-10|\leq6 \right\}$$  and $$B= \left\{x:|x-2|>1 \right\}$$. Then 

For $$A$$, 

When $$x^2<10$$, i.e. when $$x>-\sqrt{10}$$ or when $$x<\sqrt{10}$$, we get 

$$A = \{x: 10-x^{2} \leq 6 \}$$ or $$A = \{x: x^2\geq 4\}$$  

Provided the range, we get $$A = (-\sqrt{10}, -2] \cup [2, \sqrt{10})$$

When $$x^2\geq 10$$, i.e. when $$x\leq -\sqrt{10}$$ or when $$x\geq \sqrt{10}$$, we get

$$A = \{x: x^2-10\leq 6\}$$ or $$A= \{x: x^2\leq 16\}$$

Provided the range, we get $$A= [-4, -\sqrt{10}] \cup [\sqrt{10}, 4]$$

Combining both the cases, we get the set $$A = [-4, -2] \cup [2, 4]$$

For $$B$$,

When $$x< 2$$

$$B= \{x: 2-x >1 \}$$  or  $$B= \{x: x<1 \}$$

Provided the range, we get $$B= (-\infty, 1)$$

When $$x\geq 2$$

$$B= \{x: x-2 >1 \}$$ or $$B= \{x: x>3 \}$$

Provided the range, we get $$B= (3, \infty)$$

Combining both the cases, we get the set $$B= (-\infty, 1) \cup (3, \infty)$$

These give, 

$$A\cup B = (-\infty, 1)\cup [2, \infty)$$

$$A-B = [2,3]$$

$$B-A = (-\infty, -4) \cup (-2, 1) \cup (4, \infty)$$

$$A\cap B = [-4,-2]\cup (3, 4]$$ 

Only Option C satisfies, and is the correct answer.