(A) On $Z^+$,the operation $^*$ is defined by $a ^* b = 2^{ab}$.
For commutativity:
We check if $a ^* b = b ^* a$ for all $a, b \in Z^+$.
$a ^* b = 2^{ab}$
$b ^* a = 2^{ba}$
Since $ab = ba$ for all $a, b \in Z^+$,it follows that $2^{ab} = 2^{ba}$.
Thus,$a ^* b = b ^* a$.
Therefore,the operation $^*$ is commutative.
For associativity:
We check if $(a ^* b) ^* c = a ^* (b ^* c)$ for all $a, b, c \in Z^+$.
Consider $a=1, b=2, c=3$:
$(1 ^* 2) ^* 3 = (2^{1 \times 2}) ^* 3 = 4 ^* 3 = 2^{4 \times 3} = 2^{12} = 4096$.
$1 ^* (2 ^* 3) = 1 ^* (2^{2 \times 3}) = 1 ^* 2^6 = 1 ^* 64 = 2^{1 \times 64} = 2^{64}$.
Since $2^{12} \neq 2^{64}$,the operation is not associative.