1. From point $A$ to point $B$, which are 8 km apart, a tourist and a cyclist set out simultaneously. The cyclist, who took no less than half an hour to travel from $A$ to $B$, without stopping, turned back and started moving towards point $A$, increasing his speed by $25 \%$. After 10 minutes from his departure from point $B$, the cyclist met the tourist. Determine the greatest possible integer value of the tourist's speed (in km/h), and for this value of the tourist's speed, find the initial speed of the cyclist.
Solution: Let $x$ (km/h) be the speed of the tourist, $y$ (km/h) be the initial speed of the cyclist, and $t$ (h) be the time spent by the cyclist traveling from $A$ to $B$. Then

$$
\left\{\begin{array}{c}
x(t+1 / 6)+5 y / 24=8, \\
y t=8, \\
t \geq 0.5,
\end{array} \Rightarrow x(8 / y+1 / 6)+5 y / 24=8, \Rightarrow 5 y^{2}+(4 x-192) y+192 x=0\right.
$$

For the quadratic equation to have a solution, it is necessary that $D / 4=(2 x-96)^{2}-960 x \geq 0$. Therefore, $x^{2}-336 x+2304 \geq 0, D / 4=(72 \sqrt{5})^{2}$, $x \in(-\infty ; 168-72 \sqrt{5}] \cup[168+72 \sqrt{5} ;+\infty)$. Since by the condition $x \in \mathbb{N}$, and $x / 6<8$, i.e., $x<48$, then $x \in[1 ; 168-72 \sqrt{5}] \cap \mathbb{N}$. Using the estimate $2.23<\sqrt{5}<2.24$, we get the estimate $160<72 \sqrt{5}<161$, and $7<168-72 \sqrt{5}<8$. The largest possible integer value of the speed $x=7$. We find the initial speed of the cyclist when $x=7$ from the equation $5 y^{2}-164 y+192 \cdot 7=0, y_{1}=84 / 5$, $y_{2}=16$. Since $t \geq 0.5, \quad t=\frac{8}{y} \geq \frac{1}{2}$, and $y \leq 16$, then $y=16$.

Answer: 7 km/h, 16 km/h.