The term in question, $1+198/3,$ comes from the observation that to count the odd numbers $$1, 3, \dots 399$$ that are divisible by 3, the author discards 1 (obviously not divisible by 3) and 3 (obviously divisible by 3), leaving 198 odd numbers. They are all of the form $2n+3$ for $n = 1, 2, \dots, 198$ exactly one-third of these, namely when $n=3, 6, \dots, 198$ are divisible by 3, so the total number of odd integers in the range will be 1 (for the 3 the author omitted) plus $198/3.$
It's worth mentioning that there's a more general way of solving such problems, namely the Inclusion-Exclusion principle. In its simplest form it states that if you have two finite sets, $S_1, S_2,$ then their sizes are related by
$$\mid S_1 \cup S_2\mid = \mid S_1 \mid + \mid S_2 \mid-\mid S_1 \cap S_2 \mid$$
This is clear once one recognizes that the terms $\mid S_1\mid+\mid S_2\mid$ on the right count the elements in the intersection twice, so we correct the count by subtracting the elements in the intersection to get the total number of elements in either of $S_1$ or $S_2.$
This holds for more than two sets. In particular, for four sets we'll have
\mid S_1\cup S_2\cup S_3\cup S_4\mid &= \mid S_1 \mid+\mid S_2 \mid+\mid S_3 \mid + \mid S_4 \mid\\
&- \mid S_1\cap S_2\mid-\mid S_1\cap S_3\mid-\dots-\mid S_3\cap S_4\mid\\
&+ \mid S_1\cap S_2\cap S_3\mid+\dots+\mid S_2\cap S_3\cap S_4\mid\\
&-\mid S_1\cap S_2\cap S_3\cap S_4\mid
In other words on the right you add the sizes of the sets, subtract the sizes of the two-set intersections, add the sizes of the three-element intersections, and so on.
Now in your problem, we have the general result that the number of integers in the range $1, \dots 400$ which are divisible by an integer $d$ will be $\lfloor 400/d\rfloor$, so if you let $S_1$ be the elements in the range divisible by 2, $S_2$ be the elements divisible by 3, $S_3$ the ones divisible by 5, and $S_4$ the ones divisible by 7 you'll have
\mid S_1\cup S_2\cup S_3\cup S_4\mid &= 200+133+80+59\\
where, for example, $\mid S_1\cap S_2\mid$ is the number of elements in the range that are divisible by both 2 and 3 (namely by 6) so $\mid S_1\cap S_2\mid = 400/6 = 66$. Finally, the left-hand term, $\mid S_1\cup S_2\cup S_3\cup S_4\mid$ will count all the numbers in the range divisible by 2 or 3 or 5 or 7, which is what the author wanted.