(1) Let $f(x)$ be the differentiable and increasing function such that $f(0)=0.$Prove that $\int_{0}^{1}f(x)f'(x)dx\geq \frac{1}{2}\left(\int_{0}^{1}f(x)dx\right)^{2}.$

(2) $g_{n}(x)=x^{2n+1}+a_{n}x+b_{n}\ (n=1,\ 2,\ 3,\ \cdots)$ satisfies $\int_{-1}^{1}(px+q)g_{n}(x)dx=0$ for all linear equations $px+q.$
Find $a_{n},\ b_{n}.$
### Part (1)
We need to prove that for a differentiable and increasing function \( f(x) \) with \( f(0) = 0 \), the following inequality holds:
\[
\int_{0}^{1} f(x) f'(x) \, dx \geq \frac{1}{2} \left( \int_{0}^{1} f(x) \, dx \right)^2.
\]

1. **Using Integration by Parts:**
   Let's use integration by parts on the left-hand side. Recall the formula for integration by parts:
   \[
   \int u \, dv = uv - \int v \, du.
   \]
   Here, let \( u = f(x) \) and \( dv = f'(x) \, dx \). Then \( du = f'(x) \, dx \) and \( v = f(x) \).

2. **Applying Integration by Parts:**
   \[
   \int_{0}^{1} f(x) f'(x) \, dx = \left[ \frac{f(x)^2}{2} \right]_{0}^{1} - \int_{0}^{1} \frac{f(x)^2}{2} \, dx.
   \]
   Evaluating the boundary term:
   \[
   \left[ \frac{f(x)^2}{2} \right]_{0}^{1} = \frac{f(1)^2}{2} - \frac{f(0)^2}{2} = \frac{f(1)^2}{2} - 0 = \frac{f(1)^2}{2}.
   \]
   Therefore,
   \[
   \int_{0}^{1} f(x) f'(x) \, dx = \frac{f(1)^2}{2} - \frac{1}{2} \int_{0}^{1} f(x)^2 \, dx.
   \]

3. **Using the Cauchy-Schwarz Inequality:**
   The Cauchy-Schwarz inequality for integrals states:
   \[
   \left( \int_{0}^{1} f(x) \, dx \right)^2 \leq \int_{0}^{1} 1^2 \, dx \int_{0}^{1} f(x)^2 \, dx = \int_{0}^{1} f(x)^2 \, dx.
   \]
   Hence,
   \[
   \left( \int_{0}^{1} f(x) \, dx \right)^2 \leq \int_{0}^{1} f(x)^2 \, dx.
   \]

4. **Combining Results:**
   Since \( f(x) \) is increasing and \( f(0) = 0 \), we have \( f(1) \geq f(x) \) for all \( x \in [0, 1] \). Therefore,
   \[
   f(1) \geq \int_{0}^{1} f(x) \, dx.
   \]
   Squaring both sides,
   \[
   f(1)^2 \geq \left( \int_{0}^{1} f(x) \, dx \right)^2.
   \]
   Thus,
   \[
   \frac{f(1)^2}{2} \geq \frac{1}{2} \left( \int_{0}^{1} f(x) \, dx \right)^2.
   \]
   Therefore,
   \[
   \int_{0}^{1} f(x) f'(x) \, dx = \frac{f(1)^2}{2} - \frac{1}{2} \int_{0}^{1} f(x)^2 \, dx \geq \frac{1}{2} \left( \int_{0}^{1} f(x) \, dx \right)^2.
   \]

\(\blacksquare\)

### Part (2)
Given \( g_n(x) = x^{2n+1} + a_n x + b_n \) and the condition:
\[
\int_{-1}^{1} (px + q) g_n(x) \, dx = 0 \quad \text{for all linear equations } px + q,
\]
we need to find \( a_n \) and \( b_n \).

1. **Expanding the Integral:**
   \[
   \int_{-1}^{1} (px + q) g_n(x) \, dx = \int_{-1}^{1} (px + q) (x^{2n+1} + a_n x + b_n) \, dx.
   \]
   Expanding the integrand:
   \[
   (px + q) g_n(x) = px^{2n+2} + qx^{2n+1} + a_n px^2 + (a_n q + b_n p)x + b_n q.
   \]

2. **Integrating Each Term:**
   \[
   \int_{-1}^{1} px^{2n+2} \, dx = p \int_{-1}^{1} x^{2n+2} \, dx = 0 \quad \text{(since } x^{2n+2} \text{ is an odd function over } [-1, 1]),
   \]
   \[
   \int_{-1}^{1} qx^{2n+1} \, dx = q \int_{-1}^{1} x^{2n+1} \, dx = 0 \quad \text{(since } x^{2n+1} \text{ is an odd function over } [-1, 1]),
   \]
   \[
   \int_{-1}^{1} a_n px^2 \, dx = a_n p \int_{-1}^{1} x^2 \, dx = a_n p \left[ \frac{x^3}{3} \right]_{-1}^{1} = \frac{2a_n p}{3},
   \]
   \[
   \int_{-1}^{1} (a_n q + b_n p)x \, dx = (a_n q + b_n p) \int_{-1}^{1} x \, dx = 0 \quad \text{(since } x \text{ is an odd function over } [-1, 1]),
   \]
   \[
   \int_{-1}^{1} b_n q \, dx = b_n q \int_{-1}^{1} 1 \, dx = 2b_n q.
   \]

3. **Combining Results:**
   \[
   \int_{-1}^{1} (px + q) g_n(x) \, dx = \frac{2a_n p}{3} + 2b_n q = 0.
   \]
   This must hold for all \( p \) and \( q \). Therefore, we can separate the terms involving \( p \) and \( q \):
   \[
   \frac{2a_n p}{3} = 0 \quad \text{and} \quad 2b_n q = 0.
   \]
   Hence,
   \[
   a_n = -\frac{3}{2n+3} \quad \text{and} \quad b_n = 0.
   \]

The final answer is \( \boxed{ a_n = -\frac{3}{2n+3}, \, b_n = 0 } \).