Let $f(x)=\int_0^{x} e^{t} (\cos t+\sin t)\ dt,\ g(x)=\int_0^{x} e^{t} (\cos t-\sin t)\ dt.$ 

For a real number $a$, find $\sum_{n=1}^{\infty} \frac{e^{2a}}{\{f^{(n)}(a)\}^2+\{g^{(n)}(a)\}^2}.$
1. **Compute the first derivatives using the Fundamental Theorem of Calculus:**
   \[
   f'(x) = e^x (\cos x + \sin x)
   \]
   \[
   g'(x) = e^x (\cos x - \sin x)
   \]

2. **Express the derivatives using complex numbers:**
   \[
   f'(x) = e^x \left( \frac{e^{ix} + e^{-ix}}{2} + \frac{e^{ix} - e^{-ix}}{2i} \right) = \frac{1}{2} \left((1-i)e^{(1+i)x} + (1+i)e^{(1-i)x}\right)
   \]
   \[
   g'(x) = e^x \left( \frac{e^{ix} + e^{-ix}}{2} - \frac{e^{ix} - e^{-ix}}{2i} \right) = \frac{1}{2} \left((1+i)e^{(1+i)x} + (1-i)e^{(1-i)x}\right)
   \]

3. **Compute higher-order derivatives:**
   \[
   f^{(n)}(x) = \frac{1}{2} \left((1-i)(1+i)^{n-1} e^{(1+i)x} + (1+i)(1-i)^{n-1} e^{(1-i)x}\right)
   \]
   Simplifying, we get:
   \[
   f^{(n)}(x) = (1+i)^{n-2} e^{(1+i)x} + (1-i)^{n-2} e^{(1-i)x}
   \]

   Similarly,
   \[
   g^{(n)}(x) = \frac{1}{2} \left((1+i)^n e^{(1+i)x} + (1-i)^n e^{(1-i)x}\right)
   \]
   Simplifying, we get:
   \[
   g^{(n)}(x) = i \left((1+i)^{n-2} e^{(1+i)x} - (1-i)^{n-2} e^{(1-i)x}\right)
   \]

4. **Let \( u = (1+i)^{n-2} e^{(1+i)a} \) and \( v = (1-i)^{n-2} e^{(1-i)a} \):**
   \[
   f^{(n)}(a) = u + v
   \]
   \[
   g^{(n)}(a) = i(u - v)
   \]

5. **Compute the sum of squares:**
   \[
   \left(f^{(n)}(a)\right)^2 + \left(g^{(n)}(a)\right)^2 = (u+v)^2 + (i(u-v))^2
   \]
   \[
   = u^2 + 2uv + v^2 - (u^2 - 2uv + v^2)
   \]
   \[
   = 4uv
   \]
   \[
   = 4 \cdot (1+i)^{n-2} e^{(1+i)a} \cdot (1-i)^{n-2} e^{(1-i)a}
   \]
   \[
   = 4 \cdot (1+i)^{n-2} (1-i)^{n-2} e^{2a}
   \]
   \[
   = 4 \cdot (1^2 - i^2)^{n-2} e^{2a}
   \]
   \[
   = 4 \cdot (1 + 1)^{n-2} e^{2a}
   \]
   \[
   = 4 \cdot 2^{n-2} e^{2a}
   \]
   \[
   = 2^n e^{2a}
   \]

6. **Sum the series:**
   \[
   \sum_{n=1}^\infty \frac{e^{2a}}{2^n e^{2a}} = \sum_{n=1}^\infty \frac{1}{2^n}
   \]
   This is a geometric series with the first term \( \frac{1}{2} \) and common ratio \( \frac{1}{2} \):
   \[
   \sum_{n=1}^\infty \frac{1}{2^n} = \frac{\frac{1}{2}}{1 - \frac{1}{2}} = 1
   \]

The final answer is \(\boxed{1}\)