My attempt:
- $R \circ R = R^{2}$
Proving reflexively: since $R$ is symmetrical then $aRb$ and $bRa$ so $R^{2}$ will be $aR^{2}a$
Proving symmetrical: since $R$ is transitive $aRb ,bRc = aRc$ also $R$ is symmetric so $bRa,cRb = cRa$
$aRc\circ bRa = bR^{2}c$
$cRa \circ aRb = cR^{2}b $
I don't know how to prove $R^{2}$ is transitive also is my method for proving reflexive and symmetric correct?
The reflexive property of $R^2$ comes from the very same reflexive property of $R$, since when using the symmetrical property of $R$ you had to assume that such $b$ existed for $a$ in the first place, and the only $b$ that's guaranteed to exist is $a$ itself. So there's no need of the symmetrical property.
For the symmetrical property of $R^2$ I'm not sure about the meaning of what you wrote. You have to show that if $aR^2b$ then $bR^2a$. The $aR^2b$ it's the starting point, not part of the proof. Maybe I'm just a little confused. Hint for this part: Use the definition of $R^2$ for $aR^2b$ and then use the symmetrical property of $R$.
Lastly, for the transitive property of $R^2$ you just have to use the definition of $R^2$ for both $aR^2b$ and $bR^2c$ and then use the transitive property of $R$ to get $aR^2c$.