To find the value of $$|(\vec{b} \times \vec{a})-\vec{b}|^2$$, we can use properties of vector operations and magnitudes. Given $$|\vec{b}| = 1$$ and $$|\vec{b} \times \vec{a}| = 2$$, let's break down the calculation step by step:

Firstly, we observe that the cross product of two vectors $$\vec{b} \times \vec{a}$$ is orthogonal (perpendicular) to both $$\vec{b}$$ and $$\vec{a}$$. This means that when we take the dot product of $$\vec{b} \times \vec{a}$$ with either $$\vec{b}$$ or $$\vec{a}$$, the result will be zero due to the orthogonal property. Specifically,

$$(\vec{b} \times \vec{a}) \cdot \vec{b}= 0 $$(because $$\vec{b} \times \vec{a}$$ is perpendicular to both $$\vec{b}$$ and $$\vec{a}$$)

Next, to find the magnitude squared of the vector $$(\vec{b} \times \vec{a}) - \vec{b}$$, we apply the formula for the magnitude squared of a vector subtraction, which can be expressed as:

$$|(\vec{b} \times \vec{a}) - \vec{b}|^2 = |\vec{b} \times \vec{a}|^2 + |-\vec{b}|^2 + 2(\vec{b} \times \vec{a}) \cdot (-\vec{b})$$

Since $$(\vec{b} \times \vec{a}) \cdot \vec{b} = 0$$ and $$|-\vec{b}| = |\vec{b}|$$, the equation simplifies to:

$$= |\vec{b} \times \vec{a}|^2 + |\vec{b}|^2$$

Given that $$|\vec{b} \times \vec{a}| = 2$$ and $$|\vec{b}| = 1$$, substituting these values gives:

$$= 2^2 + 1^2 = 4 + 1 = 5$$

Therefore, the value of $$|(\vec{b} \times \vec{a})-\vec{b}|^2$$ is 5.