Assume that $(S,*)$ and $(T,\circ)$ are isomorphic binary structures.
(a) Show that $(S,*)$ is commutative if and only if $(T,\circ)$ is commutative.
(b) Show that $(S,*)$ is associative if and only if $(T,\circ)$ is associative
Approach: I think I know how to approach this, but just in one direction a) First direction if $(S,*)$ is commutative then $(T,\circ)$ is commutative.
Let $a,b \in S$, so $f(a*b)=f(b*a)$. $f(a*b)=f(a)\circ f(b)$ and $f(b*a)=f(b)\circ f(b)$, so $f(a)\circ f(b) = f(b)\circ f(a)$. Therefore, $(T, \circ)$ is commutative.
b) First direction if $(S, *)$ is associative then $(T, \circ)$ is associative
Let $a,b,c \in S$, so $f((a*b)*c)=f(a*(b*c))$. $f((a*b)*c) = f(a*b)\circ f(c)=(f(a)\circ f(b))\circ f(c)$ and $f(a*(b*c))=f(a)\circ f(b*c) = f(a)\circ (f(b)\circ f(c))$. This implies $(T,\circ)$ is associative.
I think that's the way to go. How does that look?
For the other directions I think that we have to make use of the fact that $f$ is bijective, so it has an inverse, namely $f^{-1} : T \to S$, but I don't know if that $f$ is homomorphic. How do I use it to prove the second direction?
For part a) you've correctly deduced that $f(a)\circ f(b) = f(b)\circ f(a)$ for all $a, b \in S$, but you haven't explained why that implies $(T, \circ)$ is commutative. That is, given $x, y \in T$, why is $x\circ y = y\circ x$? Here you need to use one of the properties of $f$. You have a similar problem for part b) which can be rectified in the same way.
As for the reverse direction, $f^{-1} : T \to S$ is a homomorphism, that's part of the definition of an isomorphism.