1) Consider the mapping p: Z3 --> Z6 given by p(x) = 2x, for x = 0,1,2. Is p a ring homomorphism? How about the mapping p(x) = remainder of 4x (mod 6)?
You first need to consider if $\displaystyle [x]_3\mapsto [2x]_6$ is well-defined. Say that $\displaystyle [x]_3 = [y]_3$ then $\displaystyle x\equiv y(\bmod 3)$, so $\displaystyle 2x\equiv 2y(\bmod 6)$ and hence $\displaystyle [2x]_6 = [2y]_6$ which means the mapping is well-defined. Now what about $\displaystyle [x]_3\mapsto [4x]_3$? Try to argue that this is also well-defined. To show it is a ring homorphism (depending on how you define "ring homomorphism") you just need to show $\displaystyle f(x+y) = f(x)f(y)$ and $\displaystyle f(xy) = f(x)f(y)$. Notice $\displaystyle f( [x]_3) + f([y]_3)= [2x]_6 + [2y]_6 = [2(x+y)]_6 = f([x+y]_3)$. Also, $\displaystyle f([x]_3)f([y]_3) = [2x]_6\cdot [2y]_6 = [4xy]_6 \not = [2xy]_6 = f([xy]_3)$. Thus, it is not a homorphism, you try doing the second case.