Can someone please explain this solution , why |a-2| can take only two values 1 and 0 ?

Because

|x² − 2|x|| ≥ 0 always

So the equation is

|x² − 2|x|| + |a − 2| = 0

|x² − 2|x|| + |a − 2| = 0

Now a sum of two non negative terms can be 0 only if both are 0

So we must have

|x² − 2|x|| = 0 and |a − 2| = 0

That gives directly

|a − 2| = 0 ⇒ a = 2

Now where does “1” come from?

That’s from the next step where they rewrite:

⇒ |x² − 2|x|| = −|a − 2|

But RHS ≤ 0 and LHS ≥ 0 ⇒ both must be 0

Then they check the borderline case where

|x² − 2|x|| can take minimum values like 0 or 1 depending on x (since it becomes (|x|−1)² type)

So effectively |a − 2| must match those possible small values, i.e. 0 or 1

That’s why

|a − 2| = 0 or 1

⇒ a = 2 or a = 1 or 3

So it’s not random, it comes from matching minimum possible values of the other non negative expression.