What would be the cardinality of $S$? Let $S$ be the set of polynomials $f(x)$ with integer coefficients satisfying $$f(x)\equiv 1\mod (x-1)$$ and $$f(x)\equiv 0\mod (x-3)$$

What would be the cardinality of $S$?

Second equation means $x=3$ is a root of $f(x)$. That means $f(3)\equiv 1\mod2 $
 and that means $0\equiv 1\mod 2$ which is a contradiction. So $S$ is empty.

Please have a look at the work. Am I right?

Thanks.
Ok, after the very last comment below the question by the OP I think the following can be what was intended: it is given $\;f(x)\in\Bbb Z[x]\;$ s.t. both the following are true:

$$\begin{cases}I\;\;\;f(x)\cong1\mod{(x-1)}\iff f(x)=1+(x-1)p(x)\;,\;\;p(x)\in\Bbb Z[x]\\{}\\
II\;\;f(x)\cong0\mod{(x-3)}\iff f(x)=(x-3)q(x)\;,\;\;q(x)\in\Bbb Z[x]\end{cases}$$

but then we get:

$$\begin{cases}I\;\implies\; f(1)=1=1\pmod 2\\{}\\
II\implies f(1)=-2q(1)=0\pmod2\end{cases}\;\;\;\implies\text{ contradiction}$$

and what the OP did is thus correct: we get that in fact $\;S=\emptyset\;$ .
$f(x)\equiv  1\bmod (x-1)$ is a congruent condition as polynomials, i.e. $f(x)=1+p(x)(x-1)$ for some polynomial $p(x)\in\mathbb{R}[x]$. In particular, $f(3)=1+p(3)(2)$. Note that $p(3)$ (a priori) does not need to be integer, so saying $f(3)\equiv 1\bmod 2$ is not justified unless you can prove that $p(3)\in\mathbb{Z}$ (which would be true if you can prove $p(x)\in\mathbb{Z}[x]$)