Find a polynomial $p(x)$ of degree $n$, so that $p(x)\ne 0$ if we plug in $x$ from $\{0,1,2,..,p-1\}$, using addition and multiplication $\mod p$. 
Find a polynomial $p(x)$ of degree $n$, so that $p(x)$ is never equal to $0$ if we plug in $x$ from $\{0,1,2,..,p-1\}$, using addition and multiplication $\mod p$.

My answer :
If make $p(x) = x(x-1)(x-2)...(x-(p-1)+1$ doesn't it satisfy the condition in the question?
 A: If the coefficients of the polynomial can be arbitrary numbers then
$$px^n+1$$ is such a polynomial of degree $n$. If we allow only coefficients from $0,\ldots,p-1$ it is not so simple. For $n=1$ and $p$ is a prime such a polynomial does not exist, because $$kx+d\equiv0 \pmod p$$ can be multiplied by the modular inverse of $k$ to get  $$x+a\equiv0 \pmod p$$ for   $x\equiv -a\pmod p$ .
if $p$ is not a prime but $p=uv, u,v>1$ then
$$ux+d\not\equiv 0 \pmod p$$
if $\gcd(u,d)=1$.
For $n=2$ and $p>2$ at least half of the numbers $1,\ldots,p-1$ do not have a square root $\pmod p$. So if $a$ has no square root such a polynomial exists: $$x^2-a$$ will not become $0 \pmod p$. This idea can be applied to arbitrary $n>1$:
So assume $p>1$  and $n>1$. For the polynomial
$$f(x)=x^n-x$$
the values $$f(0),f(1),\ldots,f(p-1)$$ are all from $\{0,\ldots,p-1\}$ but they are not all pairwise different $\pmod p$, because $$f(0)\equiv f(1)\equiv0\pmod p$$
So there is a value $$a \in \{0,\ldots,p-1\}\setminus \{f(0),f(1),\ldots,f(p-1)\}$$
Then expression
$$x^n-x-a$$
is a polynomial for $n>1$ and cannot become $0\pmod p$ for an $x$, if $p>1$.
