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.
 A: 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 $\;S=\emptyset\;$ .
A: Your argument is correct, but mixing the meaning of $\equiv$ between the ring of intergers and the ring of integer polynomials is a dangerous thing to do, because it is not obvious if one equation interpreted as over one ring will hold over the other, or even what interpretation you are using at any given moment. 
For a polynomial $p$ I'll use the notion $p(x)$ when I'm talking about the element of the polynomial ring and $p[y]$ when I'm talking about the result of 'applying' $p(x)$ to the integer $y$.
But at least one direction is easy: If $a(x),b(x)$ and $q(x)$ are integer plynomials, and
$$a(x) \equiv b(x) \pmod {q(x)}$$
is true, it is also true that
$$\forall x\in \mathbb Z:a[x] \equiv b[x] \pmod {q[x]}$$
That's because in the ring of integer polynomials we know there exists a polynomial $k(x)$ with
$$a(x)-b(x) = k(x)q(x).$$
Now addition and multiplication of polynomials have been designed such that they 'mimic' the functional evaluation, so we can derive from this that
$$\forall x\in \mathbb Z: a[x]-b[x] = k[x]q[x]$$
which then implies what we wanted to proof.
You applied the polynomial ring interpretation to your equation
$$f(x)\equiv 0\mod (x-3)$$ 
to arrive at (in my notation)
$$f[3]=0$$
and used the integer ring interpretation of 
$$f(x)\equiv 1\mod (x-1)$$ 
for $x=3$ to get (in my notation)
$$f[3] \equiv 1 \mod 2,$$
which is a contradiction, as you noted. My above proof shows that you can go from the polynomial ring interpretation to the integer ring interpretation. But as I said in the beginning, to prevent errors, one should distinguish between working in one ring and the other, if possible.
A: If $f$ is a polynomial with integer coefficients, then $f(x)$ is an integer whenever $x$ is an integer. Moreover, if $a$ and $b$ are distinct integers, then $a-b$ divides $f(a) - f(b)$.
In this case, $f$ would need to satisfy $f(1) = 1$ and $f(3) = 0$, meaning that $3-1$ would have to divide $f(3) - f(1) = -1$. 
Hence there are no polynomials with those properties.
A: $f(x)\equiv  1\bmod (x-1)$ 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 $p(x)\in\mathbb{Z}[x]$)
