Define a function $f_n(x)\ (n=0,\ 1,\ 2,\ \cdots)$ by

\[f_0(x)=\sin x,\ f_{n+1}(x)=\int_0^{\frac{\pi}{2}} f_n\prime (t)\sin (x+t)dt.\]

(1) Let $f_n(x)=a_n\sin x+b_n\cos x.$ Express $a_{n+1},\ b_{n+1}$ in terms of $a_n,\ b_n.$

(2) Find $\sum_{n=0}^{\infty} f_n\left(\frac{\pi}{4}\right).$
### Part 1: Express \(a_{n+1}\) and \(b_{n+1}\) in terms of \(a_n\) and \(b_n\)

Given:
\[ f_0(x) = \sin x \]
\[ f_{n+1}(x) = \int_0^{\frac{\pi}{2}} f_n'(t) \sin(x+t) \, dt \]

Assume:
\[ f_n(x) = a_n \sin x + b_n \cos x \]

First, we need to find the derivative of \(f_n(x)\):
\[ f_n'(x) = a_n \cos x - b_n \sin x \]

Now, substitute \(f_n'(t)\) into the integral for \(f_{n+1}(x)\):
\[ f_{n+1}(x) = \int_0^{\frac{\pi}{2}} (a_n \cos t - b_n \sin t) \sin(x+t) \, dt \]

We can expand \(\sin(x+t)\) using the angle addition formula:
\[ \sin(x+t) = \sin x \cos t + \cos x \sin t \]

Thus,
\[ f_{n+1}(x) = \int_0^{\frac{\pi}{2}} (a_n \cos t - b_n \sin t) (\sin x \cos t + \cos x \sin t) \, dt \]

Distribute and separate the terms:
\[ f_{n+1}(x) = \sin x \int_0^{\frac{\pi}{2}} a_n \cos^2 t \, dt + \sin x \int_0^{\frac{\pi}{2}} -b_n \sin t \cos t \, dt + \cos x \int_0^{\frac{\pi}{2}} a_n \cos t \sin t \, dt + \cos x \int_0^{\frac{\pi}{2}} -b_n \sin^2 t \, dt \]

Evaluate each integral separately:
\[ \int_0^{\frac{\pi}{2}} \cos^2 t \, dt = \int_0^{\frac{\pi}{2}} \frac{1 + \cos(2t)}{2} \, dt = \frac{1}{2} \left[ t + \frac{\sin(2t)}{2} \right]_0^{\frac{\pi}{2}} = \frac{\pi}{4} \]
\[ \int_0^{\frac{\pi}{2}} \sin t \cos t \, dt = \int_0^{\frac{\pi}{2}} \frac{\sin(2t)}{2} \, dt = \frac{1}{2} \left[ -\frac{\cos(2t)}{2} \right]_0^{\frac{\pi}{2}} = \frac{1}{4} \]
\[ \int_0^{\frac{\pi}{2}} \sin^2 t \, dt = \int_0^{\frac{\pi}{2}} \frac{1 - \cos(2t)}{2} \, dt = \frac{1}{2} \left[ t - \frac{\sin(2t)}{2} \right]_0^{\frac{\pi}{2}} = \frac{\pi}{4} \]

Substitute these results back into the expression for \(f_{n+1}(x)\):
\[ f_{n+1}(x) = \sin x \left( a_n \frac{\pi}{4} - b_n \frac{1}{4} \right) + \cos x \left( a_n \frac{1}{4} - b_n \frac{\pi}{4} \right) \]

Thus, we have:
\[ a_{n+1} = \frac{\pi}{4} a_n - \frac{1}{4} b_n \]
\[ b_{n+1} = \frac{1}{4} a_n - \frac{\pi}{4} b_n \]

### Part 2: Find \(\sum_{n=0}^{\infty} f_n\left(\frac{\pi}{4}\right)\)

First, note that:
\[ f_n\left(\frac{\pi}{4}\right) = a_n \sin\left(\frac{\pi}{4}\right) + b_n \cos\left(\frac{\pi}{4}\right) \]
Since \(\sin\left(\frac{\pi}{4}\right) = \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}\), we have:
\[ f_n\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} (a_n + b_n) \]

Let \(c_n = a_n + b_n\). Then:
\[ c_{n+1} = \left(\frac{\pi}{4} + \frac{1}{4}\right) a_n - \left(\frac{1}{4} + \frac{\pi}{4}\right) b_n = \left(\frac{\pi}{4} + \frac{1}{4}\right) (a_n - b_n) \]

Since \(a_n = b_n\) for all \(n\), we have:
\[ c_{n+1} = 0 \]

Thus, the series \(\sum_{n=0}^{\infty} f_n\left(\frac{\pi}{4}\right)\) converges to:
\[ \sum_{n=0}^{\infty} \frac{\sqrt{2}}{2} c_n = \frac{\sqrt{2}}{2} \sum_{n=0}^{\infty} 0 = 0 \]

The final answer is \(\boxed{0}\).