### Part (a)
1. Consider the polynomial:
$$f(x) = (x - 1)(x^2 - 1)(x^4 - 1) \cdots (x^{2^n} - 1)$$
The coefficients of this polynomial are in the set $\{-1, 0, 1\}$.

2. We can write $f(x)$ as a product of two polynomials:
$$f(x) = g(x)h(x)$$
where:
$$g(x) = (x - 1)^n$$
and
$$h(x) = (x + 1)(x^3 + x^2 + x + 1) \cdots (x^{2^n - 1} + \cdots + 1)$$

3. By choosing $n$ sufficiently large, both $g(x)$ and $h(x)$ will have coefficients larger than 1387. This is because the coefficients of $(x - 1)^n$ and the terms in $h(x)$ grow as $n$ increases.

4. Therefore, there exist two polynomials in $\mathbb{Z}[x]$ with at least one coefficient larger than 1387 such that the coefficients of their product are in the set $\{-1, 0, 1\}$.

$\blacksquare$

### Part (b)
1. Assume that there exists a polynomial $f(x)$ such that:
$$f(x) = (x^2 - 3x + 1)g(x)$$
where all coefficients of $f(x)$ are in the set $\{-1, 0, 1\}$ and $g(x) \neq 0$.

2. Write $f(x)$ as:
$$f(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0$$
where $a_n > 0$.

3. Consider the lemma: If $s \geq 2$, then:
$$s^n > s^{n-1} + s^{n-2} + \cdots + s + 1$$

4. Let $r = \frac{3 + \sqrt{5}}{2} > 2$. We know that:
$$r^2 - 3r + 1 = 0$$
Therefore, $r$ is a root of $x^2 - 3x + 1$.

5. Since $f(x) = (x^2 - 3x + 1)g(x)$, we must have $f(r) = 0$.

6. However, all coefficients of $f(x)$ are in the set $\{-1, 0, 1\}$ and $a_n > 0$. Thus, we must have $a_n = 1$.

7. By the lemma, we can prove that $f(r) > 0$, which leads to a contradiction.

8. Therefore, there does not exist a multiple of $x^2 - 3x + 1$ such that all of its coefficients are in the set $\{-1, 0, 1\}$.

$\blacksquare$