$f(x) \equiv 1$ mod $(x-1)$ and $f(x) \equiv 0$ mod $(x-3)$ then is there any $f(x)$? Let $S$ be the set of polynomials $f(x)$ with integer coefficients satisfying 
$f(x) \equiv 1$ mod $(x-1)$
$f(x) \equiv 0$ mod $(x-3)$
Which of the following statements are true?
a) $S$ is empty .
b) $S$ is  a singleton.
c)$S$ is  a finite non-empty set.
d) $S$ is  countably infinite.
My Try: I took $x =5$ then $f(5) \equiv 1$ mod $4$ and $f(5) \equiv 0$ mod $2$ . Which is impossible so $S$ is empty.
Am I correct? Is there any formal way to solve this?

We have that
$$f(x)\equiv 1\bmod{(x-1)}\iff \exists a(x)|f(x)=(x-1)a(x)+1.$$
$$f(x)\equiv 0\bmod{(x-3)}\iff \exists b(x)|f(x)=(x-3)b(x).$$
So, if $f$ exists then it must be 
$$(x-1)a(x)+1=(x-3)b(x).$$ We have for $x=1:$
$$1=-2b(1)\implies b(1)=-\dfrac 12,$$ which is not possible, since $b(x)$ is a polynomial with integer coefficientes and $1\in\mathbb{Z}$. (Note taht in a similar way we have $a(3)=-\frac12$.) Thus we conclude that there is no a polynomial with integer coefficients satisfying the given conditions. 
As it was said in comments your proof is essentially correct. If $f$ exists then it must be 
$$f(5)=4a(5)+1\equiv 1\bmod{4}$$ and $$f(5)=2b(5)\equiv 0\bmod{2}.$$ Since both congruences can't hold together you can conclude that the polynomial doesn't exist.
