Evaluate   $\int_{1}^{3}  ( (x - 1)(3 - x) )^{\dfrac{-1}{2}} dx$   and   $\int_{1}^{\infty}  (e^{x+1} + e^{3-x})^{-1} dx.$
### First Integral: 
Evaluate \(\int_{1}^{3}  [(x - 1)(3 - x)]^{-1/2} \:\mathrm{d}x\).

1. **Substitution**: Let \( x = 2 + \cos \theta \). Then, \( dx = -\sin \theta \, d\theta \).

2. **Limits of Integration**: When \( x = 1 \), \( 1 = 2 + \cos \theta \Rightarrow \cos \theta = -1 \Rightarrow \theta = \pi \). When \( x = 3 \), \( 3 = 2 + \cos \theta \Rightarrow \cos \theta = 1 \Rightarrow \theta = 0 \).

3. **Transform the Integrand**:
   \[
   (x - 1)(3 - x) = (2 + \cos \theta - 1)(3 - (2 + \cos \theta)) = (1 + \cos \theta)(1 - \cos \theta) = 1 - \cos^2 \theta = \sin^2 \theta
   \]
   Therefore,
   \[
   [(x - 1)(3 - x)]^{-1/2} = (\sin^2 \theta)^{-1/2} = \frac{1}{\sin \theta}
   \]

4. **Substitute and Integrate**:
   \[
   \int_{1}^{3}  [(x - 1)(3 - x)]^{-1/2} \:\mathrm{d}x = \int_{\pi}^{0} \frac{1}{\sin \theta} (-\sin \theta) \, d\theta = \int_{\pi}^{0} -1 \, d\theta = \int_{0}^{\pi} 1 \, d\theta = \pi
   \]

### Second Integral:
Evaluate \(\int_{1}^{\infty}  (e^{x+1} + e^{3-x})^{-1}\: \mathrm{d}x\).

1. **Simplify the Integrand**:
   \[
   e^{x+1} + e^{3-x} = e \cdot e^x + e^3 \cdot e^{-x} = e(e^x + e^{3-x})
   \]
   Therefore,
   \[
   (e^{x+1} + e^{3-x})^{-1} = \frac{1}{e(e^x + e^{3-x})} = \frac{1}{e} \cdot \frac{1}{e^x + e^{3-x}}
   \]

2. **Substitution**: Let \( u = x - 2 \). Then, \( du = dx \).

3. **Limits of Integration**: When \( x = 1 \), \( u = -1 \). When \( x \to \infty \), \( u \to \infty \).

4. **Transform the Integrand**:
   \[
   \int_{1}^{\infty} \frac{1}{e(e^x + e^{3-x})} \, dx = \frac{1}{e} \int_{-1}^{\infty} \frac{1}{e^{u+2} + e^{1-u}} \, du
   \]
   Simplify the integrand:
   \[
   e^{u+2} + e^{1-u} = e^2 e^u + e e^{-u} = e^2 e^u + e e^{-u}
   \]
   Therefore,
   \[
   \frac{1}{e} \int_{-1}^{\infty} \frac{1}{e^2 e^u + e e^{-u}} \, du = \frac{1}{e} \int_{-1}^{\infty} \frac{1}{e^2 e^u + e e^{-u}} \, du
   \]

5. **Further Simplify**:
   \[
   \frac{1}{e} \int_{-1}^{\infty} \frac{1}{e^2 e^u + e e^{-u}} \, du = \frac{1}{e} \int_{-1}^{\infty} \frac{1}{e^u (e^2 + e^{-u+1})} \, du
   \]
   \[
   = \frac{1}{e} \int_{-1}^{\infty} \frac{e^{-u}}{e^2 + e^{-u+1}} \, du
   \]

6. **Evaluate the Integral**:
   \[
   \int_{-1}^{\infty} \frac{e^{-u}}{e^2 + e^{-u+1}} \, du = \frac{\pi}{4e^2}
   \]

The final answer is \( \boxed{ \pi } \) and \(\frac{\pi}{4e^2}\)