The task is to find the inverse Laplace transforms $f(t)$ for each of the given $F(s)$ functions. Here are the three expressions provided:

1. $F(s) = \frac{2s + 3}{s^2 - 6s + 8}$
2. $F(s) = \frac{2s^2 + 6s}{s^3 - 4s^2 - 4s + 16}$
3. $F(s) = \frac{2s^2 + 8}{s^3 - 7s^2 + 14s - 8}$

Let's solve each one by performing partial fraction decomposition where necessary and using standard Laplace transform pairs.

1. Inverse Laplace of $F(s) = \frac{2s + 3}{s^2 - 6s + 8}$

First, factor the denominator: $s^2 - 6s + 8 = (s - 4)(s - 2)$ Now, apply partial fraction decomposition: $\frac{2s + 3}{(s - 4)(s - 2)} = \frac{A}{s - 4} + \frac{B}{s - 2}$ Solving for $A$ and $B$, we can find the inverse Laplace.

2. Inverse Laplace of $F(s) = \frac{2s^2 + 6s}{s^3 - 4s^2 - 4s + 16}$

This is a more complex expression, and we will need to check if we can factor or simplify the denominator, followed by decomposing into simpler fractions.

3. Inverse Laplace of $F(s) = \frac{2s^2 + 8}{s^3 - 7s^2 + 14s - 8}$

Similarly, this expression involves a cubic denominator, and factoring or simplifying the denominator will help in breaking it down into simpler terms.

I'll start solving these step by step. Let's begin with the first one and calculate the partial fractions.The partial fraction decomposition of the first function is:

$\frac{2s + 3}{(s - 4)(s - 2)} = \frac{-7}{2(s - 2)} + \frac{11}{2(s - 4)}$

Now, using the standard Laplace transform pairs:

$\mathcal{L}^{-1}\left(\frac{1}{s - a}\right) = e^{at}$

We can write the inverse Laplace transform of $F(s)$ as:

[ f(t) = -\frac{7}{2}e^{2