Given that * be the binary operation defined on $Q$.

(i) $a * b=a-b, \forall a, b \in Q$ and $b * a=b-a$

So, $$a * b \neq b * a\quad$$ $$[\because b-a \neq a-b]$$

Hence, * is not commutative.

(ii)

$$\begin{aligned} & a * b=a^2+b^2 \\ & b * a=b^2+a^2 \end{aligned}$$

So, * is commutative.

[since, ' + ' is on rational is commutative]

$$\begin{aligned} &\text { (iii) }\\ &\begin{aligned} a * b & =a+a b \\ b * a & =b+a b \\ \text { Clearly, } \quad a+a b & \neq b+a b \end{aligned}\\ &\text { So, * is not commutative. } \end{aligned}$$

(iv)

$$\begin{aligned} a * b & =(a-b)^2, \forall a, b \in Q \\ b * a & =(b-a)^2 \\ \because \quad(a-b)^2 & =(b-a)^2 \end{aligned}$$

Hence, * is commutative.

$$\begin{aligned} &\text { (i) Given that, }\\ &\begin{aligned} & a * b=1+a b, \forall a, b \in R \\ & a * b=a b+1=b * a \end{aligned} \end{aligned}$$

So, * is a commutative binary operation.

Also,

$$\begin{aligned} a *(b * c) & =a *(1+b c)=1+a(1+b c) \\ a *(b * c) & =1+a+a b c \quad\text{.... (i)}\\ (a * b) * c & =(1+a b) * c \\ & =1+(1+a b) c=1+c+a b c\quad\text{.... (ii)} \end{aligned}$$

So, * is not associative.

Hence, * is commutative but not associative.