Given the polynomials $P(x)=px^4+qx^3+rx^2+sx+t,\ Q(x)=\frac{d}{dx}P(x)$, find the real numbers $p,\ q,\ r,\ s,\ t$ such that $P(\sqrt{-5})=0,\ Q(\sqrt{-2})=0$ and $\int_0^1 P(x)dx=-\frac{52}{5}.$
1. **Given Conditions and Polynomial Forms:**
   - We are given the polynomial \( P(x) = px^4 + qx^3 + rx^2 + sx + t \).
   - The derivative of \( P(x) \) is \( Q(x) = \frac{d}{dx} P(x) \).
   - We need to find the real numbers \( p, q, r, s, t \) such that:
     - \( P(\sqrt{-5}) = 0 \)
     - \( Q(\sqrt{-2}) = 0 \)
     - \( \int_0^1 P(x) \, dx = -\frac{52}{5} \)

2. **Formulating \( P(x) \) and \( Q(x) \):**
   - Since \( P(x) \) has a root at \( \sqrt{-5} \), we can write \( P(x) \) as:
     \[
     P(x) = (x^2 + 5)(px^2 + a)
     \]
   - Expanding this, we get:
     \[
     P(x) = px^4 + 5px^2 + ax^2 + 5a = px^4 + (5p + a)x^2 + 5a
     \]
   - Comparing with the general form \( P(x) = px^4 + qx^3 + rx^2 + sx + t \), we see:
     \[
     q = 0, \quad s = 0, \quad r = 5p + a, \quad t = 5a
     \]

3. **Formulating \( Q(x) \):**
   - Differentiating \( P(x) \), we get:
     \[
     Q(x) = \frac{d}{dx} P(x) = 4px^3 + 2(5p + a)x
     \]
   - Since \( Q(x) \) has a root at \( \sqrt{-2} \), we can write:
     \[
     Q(x) = (x^2 + 2)(4px + b)
     \]
   - Expanding this, we get:
     \[
     Q(x) = 4px^3 + 2(4px + b)x = 4px^3 + 2bx
     \]
   - Comparing with the derivative form \( Q(x) = 4px^3 + 2(5p + a)x \), we see:
     \[
     b = 0 \quad \text{and} \quad 2(5p + a) = 2(5p + a)
     \]
   - Hence, \( a = -p \).

4. **Substituting \( a \) and Simplifying:**
   - Substituting \( a = -p \) into the expressions for \( r \) and \( t \):
     \[
     r = 5p + a = 5p - p = 4p
     \]
     \[
     t = 5a = 5(-p) = -5p
     \]
   - Therefore, \( P(x) \) becomes:
     \[
     P(x) = px^4 + 4px^2 - 5p
     \]

5. **Integrating \( P(x) \):**
   - We need to satisfy the integral condition:
     \[
     \int_0^1 P(x) \, dx = -\frac{52}{5}
     \]
   - Integrating \( P(x) \):
     \[
     \int_0^1 (px^4 + 4px^2 - 5p) \, dx = p \int_0^1 x^4 \, dx + 4p \int_0^1 x^2 \, dx - 5p \int_0^1 1 \, dx
     \]
     \[
     = p \left[ \frac{x^5}{5} \right]_0^1 + 4p \left[ \frac{x^3}{3} \right]_0^1 - 5p \left[ x \right]_0^1
     \]
     \[
     = p \left( \frac{1}{5} \right) + 4p \left( \frac{1}{3} \right) - 5p (1)
     \]
     \[
     = p \left( \frac{1}{5} + \frac{4}{3} - 5 \right)
     \]
     \[
     = p \left( \frac{3 + 20 - 75}{15} \right)
     \]
     \[
     = p \left( \frac{-52}{15} \right)
     \]
   - Setting this equal to the given integral value:
     \[
     p \left( \frac{-52}{15} \right) = -\frac{52}{5}
     \]
     \[
     p = 3
     \]

6. **Final Values:**
   - Substituting \( p = 3 \) into the expressions for \( q, r, s, t \):
     \[
     q = 0, \quad s = 0, \quad r = 4p = 12, \quad t = -5p = -15
     \]
   - Therefore, the polynomial \( P(x) \) is:
     \[
     P(x) = 3x^4 + 12x^2 - 15
     \]

The final answer is \( \boxed{ (p, q, r, s, t) = (3, 0, 12, 0, -15) } \)