Let $f(x)=t\sin x+(1-t)\cos x\ (0\leqq t\leqq 1)$.

Find the maximum and minimum value of the following $P(t)$.

\[P(t)=\left\{\int_0^{\frac{\pi}{2}} e^x f(x) dx \right\}\left\{\int_0^{\frac{\pi}{2}} e^{-x} f(x)dx \right\}\]
1. Define the function \( f(x) = t \sin x + (1-t) \cos x \) where \( 0 \leq t \leq 1 \).
2. We need to find the maximum and minimum values of \( P(t) \) where:
   \[
   P(t) = \left( \int_0^{\frac{\pi}{2}} e^x f(x) \, dx \right) \left( \int_0^{\frac{\pi}{2}} e^{-x} f(x) \, dx \right)
   \]
3. Let:
   \[
   A = \int_0^{\frac{\pi}{2}} e^x \sin x \, dx, \quad B = \int_0^{\frac{\pi}{2}} e^x \cos x \, dx
   \]
   \[
   C = \int_0^{\frac{\pi}{2}} e^{-x} \sin x \, dx, \quad D = \int_0^{\frac{\pi}{2}} e^{-x} \cos x \, dx
   \]
4. Using integration by parts, we can find the values of \( A, B, C, \) and \( D \):
   \[
   A = \int_0^{\frac{\pi}{2}} e^x \sin x \, dx = \left[ e^x \sin x \right]_0^{\frac{\pi}{2}} - \int_0^{\frac{\pi}{2}} e^x \cos x \, dx = e^{\frac{\pi}{2}} - B
   \]
   \[
   B = \int_0^{\frac{\pi}{2}} e^x \cos x \, dx = \left[ e^x \cos x \right]_0^{\frac{\pi}{2}} + \int_0^{\frac{\pi}{2}} e^x \sin x \, dx = e^{\frac{\pi}{2}} + A
   \]
   Solving these equations, we get:
   \[
   A + B = e^{\frac{\pi}{2}}, \quad A - B = 1
   \]
   Similarly, for \( C \) and \( D \):
   \[
   C = \int_0^{\frac{\pi}{2}} e^{-x} \sin x \, dx = \left[ e^{-x} \sin x \right]_0^{\frac{\pi}{2}} + \int_0^{\frac{\pi}{2}} e^{-x} \cos x \, dx = e^{-\frac{\pi}{2}} - D
   \]
   \[
   D = \int_0^{\frac{\pi}{2}} e^{-x} \cos x \, dx = \left[ e^{-x} \cos x \right]_0^{\frac{\pi}{2}} - \int_0^{\frac{\pi}{2}} e^{-x} \sin x \, dx = e^{-\frac{\pi}{2}} + C
   \]
   Solving these equations, we get:
   \[
   C + D = 1, \quad C - D = e^{-\frac{\pi}{2}}
   \]
5. Now, we express \( P(t) \) in terms of \( A, B, C, \) and \( D \):
   \[
   P(t) = \left[ tA + (1-t)B \right] \left[ tC + (1-t)D \right]
   \]
6. To find the critical points, we take the derivative of \( P(t) \) with respect to \( t \):
   \[
   P'(t) = \frac{d}{dt} \left[ tA + (1-t)B \right] \left[ tC + (1-t)D \right]
   \]
   Using the product rule:
   \[
   P'(t) = \left[ A - B \right] \left[ tC + (1-t)D \right] + \left[ tA + (1-t)B \right] \left[ C - D \right]
   \]
   Substituting \( A - B = 1 \) and \( C - D = e^{-\frac{\pi}{2}} \):
   \[
   P'(t) = (1) \left[ tC + (1-t)D \right] + \left[ tA + (1-t)B \right] e^{-\frac{\pi}{2}}
   \]
   Simplifying:
   \[
   P'(t) = tC + (1-t)D + tA e^{-\frac{\pi}{2}} + (1-t)B e^{-\frac{\pi}{2}}
   \]
   \[
   P'(t) = t(C + A e^{-\frac{\pi}{2}}) + (1-t)(D + B e^{-\frac{\pi}{2}})
   \]
   Setting \( P'(t) = 0 \) to find critical points:
   \[
   t(C + A e^{-\frac{\pi}{2}}) + (1-t)(D + B e^{-\frac{\pi}{2}}) = 0
   \]
   Solving for \( t \):
   \[
   t = \frac{D + B e^{-\frac{\pi}{2}}}{(D + B e^{-\frac{\pi}{2}}) - (C + A e^{-\frac{\pi}{2}})}
   \]
   Since \( A + B = e^{\frac{\pi}{2}} \) and \( C + D = 1 \), we can simplify further:
   \[
   t = \frac{1}{2}
   \]
7. Evaluating \( P(t) \) at \( t = 0, 1, \frac{1}{2} \):
   \[
   P(0) = B D, \quad P(1) = A C, \quad P\left(\frac{1}{2}\right) = \left( \frac{A + B}{2} \right) \left( \frac{C + D}{2} \right)
   \]
   \[
   P(0) = \left( e^{\frac{\pi}{2}} - 1 \right) \left( 1 - e^{-\frac{\pi}{2}} \right)
   \]
   \[
   P(1) = \left( e^{\frac{\pi}{2}} + 1 \right) \left( 1 + e^{-\frac{\pi}{2}} \right)
   \]
   \[
   P\left(\frac{1}{2}\right) = \left( \frac{e^{\frac{\pi}{2}}}{2} \right) \left( \frac{1}{2} \right) = \frac{e^{\frac{\pi}{2}}}{4}
   \]
8. Therefore, the maximum and minimum values of \( P(t) \) are:
   \[
   \boxed{P(I) = \left[ \frac{1}{2} \sinh\left(\frac{\pi}{2}\right), \frac{1}{4} e^{\frac{\pi}{2}} \right]}
   \]