Given $*:R \to R$ defined by $a*b=|a-b| \qquad a,b \in R$

$\textbf {Step 1: Checking if the operation is commutative}$:

An operation $\ast$ on $A$ is commutative if $a\ast b = b \ast a\; \forall \; a, b \in A$

$\Rightarrow a \ast b = |a-b|$

$\Rightarrow b*a=|b-a|=|-(a-b)|$ $=|a-b|$

Therefore $ a*b=b*a$. Hence operation * is commutative

$\textbf {Step 2: Checking if the operation is associative}$:

An operation $*:A \to B$ is associative if $(a*b)*c=a*(b*c) \qquad a,b,c \in A$

$\Rightarrow (a*(b*c)) = a*|b-c| = |a - |b-c||$

$\Rightarrow (a*b)*c = |a-b|*c = ||a-b|-c|$

We can show that they are not equal using a simple example, where $a=1,b=2,c=3$.

$(a*(b*c)) = |1-|2-3|| = |1-|-1|| = |1-1| = 0.$

$(a*b)*c = ||1-2| -3| = ||-1|-3| = |1-3| = |-2| = 2.$

Since $ (a*(b*c)) \neq (a*b)*c$, the * operation is not associative.

$\textbf {Step 3: Showing that a*(boc) = (aob)*(aoc)}$:

Let $a,b,c \in R$, $a*(boc)= a*(b) = |a-b|$

$(a*b) = |a-b|$ and $a*c = |a-c| \rightarrow (a*b) o (a*c) = |a-b| o |a-c| = |a-b|$

Therefore $a*(boc) = (aob)*(aoc)$:

$\textbf {Step 4: Checking to see if o distributes over *}$:

For $o$ to distribute over $\ast$, the following must be true: $ao(b*c) = (aob)*(boc)$

Consider $ao(b*c) \rightarrow ao(|b-c|) = a$

Consider $(aob) = a$ and $(boc) = b \rightarrow (aob)*(boc) = a*b = |a-b|$

Therefore, $ao(b*c) \neq (aob)*(boc) \rightarrow o$ does not distribute over $\ast$.