$f(x) \equiv 1$ 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?

Your solution is short and correct. Indeed:
$$\begin{align}\begin{cases}f(x)\equiv 1\bmod{(x-1)}\\ f(x)\equiv 0\bmod{(x-3)}\end{cases} &\iff
\begin{cases}f(x)=(x-1)a(x)+1\\ f(x)=(x-3)b(x) \end{cases} \Rightarrow \\
\begin{cases}f(5)\equiv 1\bmod{4}\\f(5)\equiv 0\bmod{2}\end{cases} &\iff\begin{cases} f(5)=4a(5)+1\\ f(5)=2b(5) \end{cases}\\
0\equiv 1\bmod 2&\iff 0=2(2a(5)-b(5))+1 \Rightarrow \\
\emptyset &\iff \emptyset
\end{align}$$
