Combining both parts of the $n$-th hour, the total distance travelled forward is $\frac{1}{n^2+1} - \frac{1}{n+1} \cdot \frac{1}{n^2+1} = \frac{n}{(n+1)(n^2+1)}$ metres. On the first hour, this means $\frac{1}{4}$ metres. Notice that $\frac{n}{(n+1)(n^2+1)} < \frac{n}{(n+1)n} = \frac{1}{n} - \frac{1}{n+1}$. Thus by the end of the $m$-th hour, where $m \ge 2$, they have travelled less than $\frac{1}{4} + (\frac{1}{2} - \frac{1}{3}) + (\frac{1}{3} - \frac{1}{4}) + \dots + (\frac{1}{m} - \frac{1}{m+1}) = \frac{3}{4} - \frac{1}{m+1}$ metres, which is less than $75$ centimetres. Thus they will never reach the neighbour by the end of the hour.

We will show that they will also not reach the neighbour between the two parts of an hour. Indeed, $45$ minutes after starting they have gone $\frac{1}{2}$ metres and $1$ hour and $45$ minutes after starting $\frac{1}{4} + \frac{1}{5}$ metres, both less than $75$ centimetres. But if $m \ge 2$, then $m$ hours and $45$ minutes after starting they have travelled less than $\frac{3}{4} - \frac{1}{m+1} + \frac{1}{(m+1)^2+1}$ metres, which is still less than $75$ centimetres, as $\frac{1}{(m+1)^2+1} < \frac{1}{m+1}$.

Thus they will never reach the neighbour.