Sure, I'll provide detailed steps for each exercise.

Exercise 1: Laplace Transforms

1. $\frac{1 - e^{-2x}}{2x}$

To find the Laplace transform of $\frac{1 - e^{-2x}}{2x}$, we use the linearity property of the Laplace transform.

$\mathcal{L}\left\{\frac{1 - e^{-2x}}{2x}\right\} = \frac{1}{2} \mathcal{L}\left\{\frac{1}{x}\right\} - \frac{1}{2} \mathcal{L}\left\{\frac{e^{-2x}}{x}\right\}$

However, note that $\frac{1}{x}$ and $\frac{e^{-2x}}{x}$ are not standard forms that have direct Laplace transforms. Instead, consider rewriting the problem using properties of exponential functions and known transforms:

Using the fact that: $\mathcal{L}\{e^{-ax}f(x)\} = F(s + a)$

Rewriting: $\mathcal{L}\left\{\frac{1 - e^{-2x}}{2x}\right\} = \frac{1}{2} \left( \mathcal{L}\left\{\frac{1}{x}\right\} - \mathcal{L}\left\{e^{-2x} \cdot \frac{1}{x}\right\} \right)$

Given that the Laplace transform of $\frac{1}{x}$ does not exist, this expression might be approached through convolution theorem or residue theorem in complex analysis which is advanced. Typically in problems, standard pairs or further simplifications are used.

2. $x \int_0^\pi e^{-x t} \cos t \, dt$

Let's denote the integral by $I(x)$:

$I(x) = \int_0^\pi e^{-xt} \cos t \, dt$

We apply integration by parts:

Let $u = \cos t$ and $dv = e^{-xt} dt$.

Then, $du = -\sin t dt$ and $v = \frac{e^{-xt}}{-x}$.

$I(x) = \left. \frac{\cos t \cdot e^{-xt}}{-x} \right|_0^\pi + \int_0^\pi \frac{\sin t \cdot e^{-xt}}{-x} dt$

Simplify each part:

$I(x) = \left( \frac{\cos \pi \cdot e^{-x\pi}}{-x} - \frac{\cos 0 \cdot e^{-0}}{-x} \right) + \frac{1}{x} \int_0^\pi \sin t \cdot e^{-xt} dt$

$I(x) = \left( \frac{-e^{-x\pi}}{-x} - \frac{1}{-x} \right) + \frac{1}{x} \int_0^\pi \sin t \cdot e^{-xt} dt$

Simplifying further:

$I(x) = \frac{e^{-x\pi} - 1}{x} + \frac{1}{x} \int_0^\pi \sin t \cdot e^{-xt} dt$

We need to solve $\int_0^\pi \sin t \cdot e^{-xt} dt$. Similarly, this involves integration by parts or lookup from integral tables.

3. $e^{-x} \int_0^x (x - t)^2 \cos^2 t \, dt$

This problem typically involves solving the integral first before applying the Laplace transform. Let’s denote the integral by $J(x)$:

$J(x) = \int_0^x (x - t)^2 \cos^2 t \, dt$

Using the trigonometric identity $\cos^2 t = \frac{1 + \cos 2t}{2}$:

$J(x) = \int_0^x (x - t)^2 \cdot \frac{1 + \cos 2t}{2} dt$

Split the integral:

[ J(x) = \frac{1}{2} \int_0^x (