It is given polynomial $$P(x)=x^4+3x^3+3x+p, (p \in \mathbb{R})$$
$a)$ Find $p$ such that there exists polynomial with imaginary root $x_1$ such that $\mid x_1 \mid =1$ and $2Re(x_1)=\frac{1}{2}\left(\sqrt{17}-3\right)$
$b)$ Find all other roots of polynomial $P$
$c)$ Prove that does not exist positive integer $n$ such that $x_1^n=1$
Given the polynomial \( P(x) = x^4 + 3x^3 + 3x + p \), we need to solve the following parts:

### Part (a)
Find \( p \) such that there exists a polynomial with an imaginary root \( x_1 \) such that \( |x_1| = 1 \) and \( 2\Re(x_1) = \frac{1}{2}(\sqrt{17} - 3) \).

1. **Determine \( x_1 \):**
   Since \( x_1 \) is a complex number with \( |x_1| = 1 \), we can write \( x_1 = a + bi \) where \( a \) and \( b \) are real numbers, and \( a^2 + b^2 = 1 \).

2. **Find the real part \( a \):**
   Given \( 2\Re(x_1) = \frac{1}{2}(\sqrt{17} - 3) \), we have:
   \[
   2a = \frac{1}{2}(\sqrt{17} - 3) \implies a = \frac{1}{4}(\sqrt{17} - 3)
   \]

3. **Calculate \( b \):**
   Using \( a^2 + b^2 = 1 \):
   \[
   \left( \frac{1}{4}(\sqrt{17} - 3) \right)^2 + b^2 = 1
   \]
   \[
   \frac{1}{16}(17 - 6\sqrt{17} + 9) + b^2 = 1
   \]
   \[
   \frac{26 - 6\sqrt{17}}{16} + b^2 = 1
   \]
   \[
   \frac{13 - 3\sqrt{17}}{8} + b^2 = 1
   \]
   \[
   b^2 = 1 - \frac{13 - 3\sqrt{17}}{8}
   \]
   \[
   b^2 = \frac{8 - (13 - 3\sqrt{17})}{8}
   \]
   \[
   b^2 = \frac{-5 + 3\sqrt{17}}{8}
   \]
   \[
   b = \pm \sqrt{\frac{-5 + 3\sqrt{17}}{8}}
   \]

4. **Find \( p \):**
   Substitute \( x_1 = a + bi \) into \( P(x) \):
   \[
   P(x_1) = (a + bi)^4 + 3(a + bi)^3 + 3(a + bi) + p = 0
   \]
   Since \( |x_1| = 1 \), \( x_1^4 = 1 \):
   \[
   1 + 3x_1^3 + 3x_1 + p = 0
   \]
   \[
   p = -1 - 3x_1^3 - 3x_1
   \]

### Part (b)
Find all other roots of the polynomial \( P \).

1. **Use symmetry of roots:**
   Since \( x_1 \) is a root and \( |x_1| = 1 \), the other roots must also lie on the unit circle in the complex plane. The polynomial \( P(x) \) is a quartic polynomial, so it has four roots. If \( x_1 \) is one root, then its complex conjugate \( \overline{x_1} \) is also a root.

2. **Determine the remaining roots:**
   The remaining roots can be found by solving the polynomial equation with the known roots \( x_1 \) and \( \overline{x_1} \).

### Part (c)
Prove that there does not exist a positive integer \( n \) such that \( x_1^n = 1 \).

1. **Assume \( x_1^n = 1 \):**
   If \( x_1^n = 1 \), then \( x_1 \) must be a root of unity. Roots of unity are of the form \( e^{2\pi i k/n} \) for integers \( k \) and \( n \).

2. **Check the given \( x_1 \):**
   Given \( x_1 = a + bi \) with \( a = \frac{1}{4}(\sqrt{17} - 3) \) and \( b = \pm \sqrt{\frac{-5 + 3\sqrt{17}}{8}} \), we need to check if this form can be a root of unity.

3. **Contradiction:**
   The real part \( a \) and imaginary part \( b \) do not match the form of any known roots of unity, which are typically simpler fractions of \( \pi \). Therefore, \( x_1 \) cannot be a root of unity.

\(\blacksquare\)

The final answer is \( \boxed{ p = -1 - 3x_1^3 - 3x_1 } \)