Consider the given expression:
Let us expand the first term:$$\Bigg(\dfrac{a}{2b}+\dfrac{2b}{a}\Bigg)^2$$
We know that
$$(a+b)^2=a^2+b^2+2ab$$
$$\therefore \Bigg(\dfrac{a}{2b}+\dfrac{2b}{a}\Bigg)^2= \Bigg(\dfrac{a}{2b}\Bigg)^2+\Bigg(\dfrac{2b}{a}\Bigg)^2+2 \times \dfrac{a}{2b} \times \dfrac{2b}{a} $$
$$\therefore \Bigg(\dfrac{a}{2b}+\dfrac{2b}{a}\Bigg)^2= \dfrac{a^2}{4b^2}+\dfrac{4b^2}{a^2}+2.....(1)$$
Let us expand the second term:$$\Bigg(\dfrac{a}{2b}-\dfrac{2b}{a}\Bigg)^2$$
We know that
$$(a-b)^2=a^2+b^2-2ab$$
$$\therefore \Bigg(\dfrac{a}{2b}+\dfrac{2b}{a}\Bigg)^2= \Bigg(\dfrac{a}{2b}\Bigg)^2+\Bigg(\dfrac{2b}{a}\Bigg)^2-2 \times \dfrac{a}{2b} \times \dfrac{2b}{a} $$
$$\therefore \Bigg(\dfrac{a}{2b}+\dfrac{2b}{a}\Bigg)^2= \dfrac{a^2}{4b^2}+\dfrac{4b^2}{a^2}-2.....(1)$$
Thus from $$(1)$$ and $$(2)$$, the given expression is
$$\Bigg(\dfrac{a}{2b}+\dfrac{2b}{a}\Bigg)^2-\Bigg(\dfrac{a}{2b}-\dfrac{2b}{a}\Bigg)^2-4=\dfrac{a^2}{4b^2}+\dfrac{4b^2}{a^2}+2-\dfrac{a^2}{4b^2}-\dfrac{4b^2}{a^2}+2-4=0$$