### Problem 1:
1. Let $a_1$ be the initial number of amoebas and $b_1$ be the initial number of bacteria.
2. On the second day, each amoeba and each bacterium divide into two cells. Thus, the number of amoebas and bacteria both double:
$$a_2 = 2a_1, \quad b_2 = 2b_1$$
3. After division, each amoeba preys on exactly one bacterium. Therefore, the number of bacteria decreases by the number of amoebas:
$$b_2' = b_2 - a_2 = 2b_1 - 2a_1$$
4. This pattern continues every day. On the $n$-th day, the number of amoebas and bacteria before predation are:
$$a_n = 2^{n-1}a_1, \quad b_n = 2^{n-1}b_1$$
5. After predation on the $n$-th day:
$$b_n' = b_n - a_n = 2^{n-1}b_1 - 2^{n-1}a_1 = 2^{n-1}(b_1 - a_1)$$
6. On the 100th day, it is given that all bacteria are killed, i.e., $b_{100}' = 0$:
$$2^{99}(b_1 - a_1) = 0$$
7. Since $2^{99} \neq 0$, it must be that:
$$b_1 - a_1 = 0 \implies b_1 = a_1$$
8. Therefore, the ratio of the number of amoebas to the number of bacteria on the first day is:
$$\boxed{1}$$

### Problem 2:
1. Given the function:
$$f(n) = \frac{4n + \sqrt{4n^2 - 1}}{\sqrt{2n+1} + \sqrt{2n-1}}$$
2. Rationalize the denominator by multiplying the numerator and the denominator by the conjugate of the denominator:
$$f(n) = \frac{(4n + \sqrt{4n^2 - 1})(\sqrt{2n+1} - \sqrt{2n-1})}{(\sqrt{2n+1} + \sqrt{2n-1})(\sqrt{2n+1} - \sqrt{2n-1})}$$
3. Simplify the denominator:
$$(\sqrt{2n+1})^2 - (\sqrt{2n-1})^2 = (2n+1) - (2n-1) = 2$$
4. The numerator becomes:
$$(4n + \sqrt{4n^2 - 1})(\sqrt{2n+1} - \sqrt{2n-1})$$
5. Simplify the expression:
$$f(n) = \frac{4n(\sqrt{2n+1} - \sqrt{2n-1}) + \sqrt{4n^2 - 1}(\sqrt{2n+1} - \sqrt{2n-1})}{2}$$
6. Notice that the terms involving the square roots will cancel out, leaving:
$$f(n) = 2n$$
7. Therefore, the sum $f(13) + f(14) + \cdots + f(112)$ is:
$$\sum_{n=13}^{112} 2n = 2 \sum_{n=13}^{112} n$$
8. The sum of the first $m$ integers is given by:
$$\sum_{k=1}^{m} k = \frac{m(m+1)}{2}$$
9. Therefore:
$$\sum_{n=13}^{112} n = \sum_{n=1}^{112} n - \sum_{n=1}^{12} n = \frac{112 \cdot 113}{2} - \frac{12 \cdot 13}{2}$$
10. Simplify:
$$\sum_{n=13}^{112} n = 6356 - 78 = 6278$$
11. Thus:
$$2 \sum_{n=13}^{112} n = 2 \cdot 6278 = 12556$$
12. Therefore:
$$\boxed{12556}$$

### Problem 3:
1. The first nine balls form a 3x3 square on the table. Each ball has a radius of 10 cm.
2. The next four balls are placed on top of the first nine balls, forming a 2x2 square.
3. The fourteenth ball is placed on top of the four balls.
4. The height of the center of the topmost ball is the sum of the heights of the layers of balls.
5. The height of the first layer (3 balls) is:
$$2 \times 10 = 20 \text{ cm}$$
6. The height of the second layer (2 balls) is:
$$2 \times 10 = 20 \text{ cm}$$
7. The height of the third layer (1 ball) is:
$$2 \times 10 = 20 \text{ cm}$$
8. Therefore, the total height is:
$$20 + 20 + 20 = 60 \text{ cm}$$
9. Therefore:
$$\boxed{60 \text{ cm}}$$

### Problem 4:
1. Given a triangle $ABC$ with sides $5$ cm, $8$ cm, and $\sqrt{41}$ cm.
2. Use Heron's formula to find the area of the triangle:
$$s = \frac{5 + 8 + \sqrt{41}}{2}$$
3. Simplify:
$$s = \frac{13 + \sqrt{41}}{2}$$
4. The area $A$ of the triangle is:
$$A = \sqrt{s(s-5)(s-8)(s-\sqrt{41})}$$
5. Simplify:
$$A = \sqrt{\left(\frac{13 + \sqrt{41}}{2}\right)\left(\frac{13 + \sqrt{41}}{2} - 5\right)\left(\frac{13 + \sqrt{41}}{2} - 8\right)\left(\frac{13 + \sqrt{41}}{2} - \sqrt{41}\right)}$$
6. The maximum possible area of the rectangle that can be inscribed in the triangle is half the area of the triangle.
7. Therefore:
$$\boxed{\frac{A}{2}}$$

### Problem 5:
1. There are 12 people in line, 5 of whom have 10,000 Rp bills and 7 have 5,000 Rp bills.
2. The ticket seller initially has 5,000 Rp.
3. The number of ways to arrange the 12 people is:
$$\frac{12!}{5!7!} = 792$$
4. The number of valid arrangements where the ticket seller has enough change is 49.
5. Therefore, the probability is:
$$\boxed{\frac{49}{792}}$$