I first show all three numbers are distinct. Assume $a=b$. Then $a\mid p$ but $p>a>1$, a contradiction. Equipped with this, part a is clear. Indeed if, say, $a^2\equiv b^2\pmod{p}$, then $p\mid (a-b)(a+b)$ and it is clear that $p>\max\{|a-b|,a+b\}$. We now turn our attention to b, which is more interesting. Assume $a^3\equiv b^3\pmod{p}$. Then $p\mid a^2+ab+b^2$. Now, note that $a(b+c)+bc\equiv 0\pmod{p}$, and thus $a\equiv -bc(b+c)^{-1}\pmod{p}$ (check that $p>b+c$ and thus $(b+c)^{-1}$ modulo $p$ exists). Inserting this and doing a bit of algebra, we get $p\mid b^2+bc+c^2$. We are now ready to conclude. Combining the facts above, we get \[ p\mid a^2+ab-bc-c^2 = (a-c)(a+b+c). \]Clearly $p\nmid a-c$ as $p>|a-c|>0$. Moreover, $p>a+b+c$ as $a,b,c>1$. A contradiction is reached.