1. Determine all irreducible fractions $\frac{a}{b}, a, b \in \mathbb{N}$, such that

$$
\frac{a}{b}-\frac{b}{a}=2 \frac{71}{80}
$$
Solution. From the condition of the problem, it follows that $\frac{a}{b}>\frac{b}{a}$, from which we get $a^{2}>b^{2}$ and since $a, b \in \mathbb{N}$, it holds that $a>b$. Furthermore, since the fraction $\frac{a}{b}$ is irreducible, we have $\operatorname{GCD}(a, b)=1$. Now, from $\frac{a}{b}-\frac{b}{a}=2 \frac{71}{80}$, it follows sequentially that

$$
\begin{aligned}
& \frac{a}{b}-\frac{b}{a}=\frac{231}{80} \\
& 80\left(a^{2}-b^{2}\right)=231 a b
\end{aligned}
$$

However, $\operatorname{GCD}(80,231)=1$, so from the last equality, it follows that $a$ and $b$ are divisors of 80. The divisors of 80 are: $1,2,4,5,8,10,16,20,40$ and 80, and by direct verification, we find that the only solution to the last equation is $a=16, b=5$. Therefore, the only irreducible fraction that satisfies the condition of the problem is $\frac{a}{b}=\frac{16}{5}$.