The digital root of a number is (almost) the remainder when the number is divided by $9$. The exception is that when a number is divisible by $9$, its digital root is traditionally taken to be $0$ or $9$. (I prefer $0$.)
Suppose for example that the digital root of $x$ is $8$. Then $x=9q+8$ for some $q$. It follows that $x^2=(9q+8)^2=81q^2+144q+64=9(9q^2+16q+7)+1$.
So the remainder when $x^2$ is divided by $9$ is $1$. It follows that the digital root of $x^2$ is $1$.
The others are done the same way.
**Remark:** We avoided congruence notation. It would make the calculation much simpler. For example, suppose that the digital root of $x$ is $4$. Then $x\equiv 4\pmod 9$. Thus $x^2\equiv 4^2\equiv 7\pmod{9}$.