2. Express the fraction $\frac{93}{91}$ as the sum of two positive fractions whose denominators are 7 and 13.
First method.

We seek $a, b \in \mathbb{N}$ such that $\frac{93}{91}=\frac{a}{7}+\frac{b}{13}=\frac{13 a+7 b}{91}$.

Thus, $93=13 a+7 b$ or $13 a=93-7 b, a, b \in \mathbb{N}$.

Since $b \geq 1$, then $7 b \geq 7$ and thus $13 a=93-7 b \leq 86$ which means $a \leq 6$.

From $93=13 a+7 b$ it follows that $7 b=93-13 a$, and since $7 b$ is divisible by 7, then $93-13 a$ must also be divisible by 7.

We have

| $a$ | 1 | 2 | 3 | 4 | 5 | 6 |
| :---: | :--- | :--- | :--- | :--- | :--- | :--- |
| $93-13 a$ | 80 | 67 | 54 | 41 | 28 | 15 |

Since only the number 28 is divisible by 7, it follows that $a=5$ and $b=4$.

The desired representation is $\frac{93}{91}=\frac{5}{7}+\frac{4}{13}$.