Solve all together and indicate the question numbers

Let A be the set of attackers, D be the set of defenders, and M be the set of midfielders.
Total players = 20.
Given:
$|A| = 8$
$|D| = 12$
$|M| = 12$
$|A \cap M| = 5$ (attack and midfield)
$|D \cap M| = 6$ (defense and midfield)
$|A \setminus (M \cup D)| = 2$ (only attack)
$|A \cap D \cap M| = 3$ (all three roles)

• Players in all three roles ($A \cap D \cap M$): 3
• Players in attack and midfield only ($A \cap M \setminus D$): $|A \cap M| - |A \cap D \cap M| = 5 - 3 = 2$
• Players in defense and midfield only ($D \cap M \setminus A$): $|D \cap M| - |A \cap D \cap M| = 6 - 3 = 3$
• Players in attack only ($A \setminus (M \cup D)$): 2
• Players in attack and defense only ($A \cap D \setminus M$): $|A \cap D \setminus M| = |A| - (|A \setminus (M \cup D)| + |A \cap M \setminus D| + |A \cap D \cap M|)$ $|A \cap D \setminus M| = 8 - (2 + 2 + 3) = 8 - 7 = 1$
• Players in defense only ($D \setminus (A \cup M)$): $|D \setminus (A \cup M)| = |D| - (|A \cap D \setminus M| + |D \cap M \setminus A| + |A \cap D \cap M|)$ $|D \setminus (A \cup M)| = 12 - (1 + 3 + 3) = 12 - 7 = 5$
• Players in midfield only ($M \setminus (A \cup D)$): $|M \setminus (A \cup D)| = |M| - (|A \cap M \setminus D| + |D \cap M \setminus A| + |A \cap D \cap M|)$ $|M \setminus (A \cup D)| = 12 - (2 + 3 + 3) = 12 - 8 = 4$

For $f(x) = 1 - 2x$:
The domain of $f$ is $\boxed{\mathbb{R}}$.

For $g(x) = \frac{1}{x+3}$:
The domain of $g$ is $\boxed{\{x \in \mathbb{R} \mid x \neq -3\}}$.

Assume $g(a) = g(b)$.

Conclusion: The function $g(x)$ is injective (or one-to-one).

Let $y = g(x) = \frac{1}{x+3}$. Swap $x$ and $y$:

The inverse function $g^{-1}(x)$ is $\boxed{\frac{1 - 3x}{x}}$.

Using $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ for $a=4, b=9, c=-3$:

The values of $x$ are $\boxed{\frac{-9 \pm \sqrt{129}}{8}}$.

Let the first term be $a$ and the common difference be $d$.
The second, fourth, and eighth terms are $a+d$, $a+3d$, and $a+7d$. These form a G.P.:

The sum of the third and fifth terms is 20:

Substitute $(2)$ into $(1)$:

This gives $d=0$ or $d=\frac{5}{2}$.
If $d=0$, then $a=10$. If $d=\frac{5}{2}$, then $a = 10 - 3\left(\frac{5}{2}\right) = 10 - \frac{15}{2} = \frac{5}{2}$.
We use $a=\frac{5}{2}$ and $d=\frac{5}{2}$ for a non-trivial sequence.

I. The first four terms

The first four terms are $\boxed{\frac{5}{2}, 5, \frac{15}{2}, 10}$.

II. The sum of the first ten terms of the arithmetic sequence

Using $S_n = \frac{n}{2}(2a + (n-1)d)$:

The sum of the first ten terms is $\boxed{\frac{275}{2} \text{ or } 137.5}$.