Explanation

First find set $A$.

$$ |(|x-3|-3)|\le1 $$

This gives

$$ -1\le |x-3|-3 \le1 $$

Adding $3$ throughout,

$$ 2\le |x-3|\le4 $$

So,

$$ |x-3|=2,3,4 $$

Corresponding integer values:

$$ |x-3|=2 \Rightarrow x=1,5 $$ $$ |x-3|=3 \Rightarrow x=0,6 $$ $$ |x-3|=4 \Rightarrow x=-1,7 $$

Hence,

$$ A=\{-1,0,1,5,6,7\},\quad |A|=6 $$

Now find set $B$.

Given

$$ \frac{(x-2)(x-4)}{x-1}\log_e(|x-2|)=0 $$

Product is zero when at least one factor is zero.

Case 1:

$$ \log_e(|x-2|)=0 $$

$$ |x-2|=1 \Rightarrow x=1,3 $$

But $x=1$ is excluded, so $x=3$ is valid.

Case 2:

$$ (x-2)(x-4)=0 \Rightarrow x=2,4 $$

$x=2$ is excluded, so $x=4$ is valid.

Thus,

$$ B=\{3,4\},\quad |B|=2 $$

Number of onto functions from a set of $6$ elements to a set of $2$ elements is

$$ 2^6-2 $$

Subtracting the two constant (non-onto) functions.

$$ =64-2=62 $$

Hence, the required number of onto functions is 62.

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