1. According to the problem, Sasha makes one step in 1 second, while Dan makes one step in $\frac{6}{7}$ seconds. Therefore, after 6 seconds, both Sasha and Dan will have made an integer number of steps, specifically, Sasha will have made 6 steps, and Dan will have made 7 steps. Consequently, we need to consider moments of time that are multiples of 6 seconds, i.e., $t=6k, k=1,2, \ldots$. Since Sasha's step length is 0.4 meters and Dan's step length is 0.6 meters, in $t=6k$ seconds, Sasha will have walked $6 \cdot k \cdot 0.4 = 2.4k$ meters, and Dan will have walked $7 \cdot k \cdot 0.6 = 4.2k$ meters. The distance between Sasha and Dan on the road at such moments of time is:

$$
d(k)=|200-2.4k|+|300-4.2k|.
$$

Consider the function $d(x), x \in \mathbb{R}$. For $x \leq x_{1}=\frac{300}{4.2}=71.42 \ldots$

$$
d(x)=200-2.4x+300-4.2x=500-6.6x
$$

For $x_{1} \leq x \leq x_{2}=\frac{200}{2.4}=83$,(3)

$$
d(x)=200-2.4x-300+4.2x=1.8x-100
$$

For $x \geq x_{2}$

$$
d(x)=-200+2.4x-300+4.2x=6.6x-500
$$

We have obtained that for $x \leq x_{1}$ the function $d(x)$ is decreasing, and for $x \geq x_{1}$ it is increasing. This means that at the point $x=x_{1}$ the function $d(x)$ attains its minimum value. To find the minimum of the function $d(k)$, consider the nearest integer values of $k$ to $x=x_{1}$: $k_{1}=\left[x_{1}\right]=71$ and $k_{2}=\left[x_{1}\right]+1=72$. Calculate $d(71)=500-6.6 \cdot 71=31.4$ and $d(72)=1.8 \cdot 72-100=29.6$. Therefore, the smallest possible distance between Sasha and Dan is 29.6 meters. Since this distance is achieved at $k=72$, the number of steps Sasha has taken by this time is $6 \cdot 72=432$, and the number of steps Dan has taken is $7 \cdot 72=504$.
Answer: 1) $d_{\min }=29.6$ m; 2) Sasha made 432 steps, Dania made 504 steps.