If $p(x)=a_{0}+a_{1} x+\cdots+a_{m} x^{m}$ is a polynomial with real coefficients $a_{i},$ then set
$$
\Gamma(p(x))=a_{0}^{2}+a_{1}^{2}+\cdots+a_{m}^{2}.
$$
Let $F(x)=3 x^{2}+7 x+2 .$ Find, with proof, a polynomial $g(x)$ with real coefficients such that

(i) $g(0)=1,$ and
(ii) $\Gamma\left(f(x)^{n}\right)=\Gamma\left(g(x)^{n}\right)$

for every integer $n \geq 1.$
1. **Identify the polynomial \( g(x) \):**
   We need \( g(x) \) to satisfy \( g(0) = 1 \). Let's consider a polynomial of the form \( g(x) = ax + b \).

2. **Calculate \( \Gamma(g(x)) \):**
   For \( g(x) = ax + b \), we have:
   \[
   \Gamma(g(x)) = a^2 + b^2
   \]

3. **Match \( \Gamma(g(x)) \) with \( \Gamma(F(x)) \):**
   We need \( \Gamma(g(x)^n) = \Gamma(F(x)^n) \). For \( n = 1 \), we have:
   \[
   \Gamma(g(x)) = \Gamma(F(x)) = 62
   \]
   Therefore, we need:
   \[
   a^2 + b^2 = 62
   \]

4. **Ensure \( g(0) = 1 \):**
   Since \( g(0) = b \), we need \( b = 1 \). Thus:
   \[
   a^2 + 1^2 = 62 \implies a^2 + 1 = 62 \implies a^2 = 61 \implies a = \sqrt{61} \text{ or } a = -\sqrt{61}
   \]

5. **Form the polynomial \( g(x) \):**
   We have two possible polynomials:
   \[
   g(x) = \sqrt{61}x + 1 \quad \text{or} \quad g(x) = -\sqrt{61}x + 1
   \]

6. **Verify the conditions:**
   - \( g(0) = 1 \) is satisfied for both polynomials.
   - We need to check if \( \Gamma(g(x)^n) = \Gamma(F(x)^n) \) for all \( n \geq 1 \). Since \( \Gamma(g(x)) = 62 \) and \( \Gamma(F(x)) = 62 \), and both polynomials have the same form, this condition is satisfied.

Therefore, the polynomial \( g(x) \) that satisfies the given conditions is:

\[
g(x) = \sqrt{61}x + 1 \quad \text{or} \quad g(x) = -\sqrt{61}x + 1
\]

The final answer is \( \boxed{ g(x) = \sqrt{61}x + 1 } \) or \( g(x) = -\sqrt{61}x + 1 \)