(No spoilers, so proceed with caution)
There are 9 solutions for the first, around 17 for the second, and many for the other two.
Recurring Heuristics
(o) Check the future "timeline" of path numbers: By "timeline", I mean "the numbers that the path can turn into". In the second puzzle, the two exit's maximum timeline is $2 \rightarrow3\rightarrow4\rightarrow5\rightarrow6\rightarrow7\rightarrow8\rightarrow9$. But as $26$ must go at second place, and take a new exit, the max timeline becomes $2 \rightarrow3\rightarrow4\rightarrow5\rightarrow6\rightarrow7\rightarrow8$.
(i) Check the order of the first and the last numbers.: For example, in the first puzzle, the beginning and the end is $(3,1)$. So in future, there would be $(3,1), (4,2), (5,3), (6,4)$. Now, $(5,3)$ must be for the $15$, and $(6,4)$ must be for the $12$. $(3,1)$ has now only one choice, so it must be $18$, so and $(4,2)$ must be $8$. This makes the order unique. Also, for example, in the second puzzle, $26$ must go for a exit for the first time.
(ii) Find upper bounds on link numbers: After a certain point, you can ignore any prime factors since they aren't possible to "reach" by th tokens. In #3, the max number is $\text{maximum node number}+\text{number of tokens} = 8+8 = 16$. For the second, the useful prime factors are $ 9 \rightarrow 1,3,9; \; \; 12 \rightarrow 1,2,3,4,6,12 \; \; 20 \rightarrow 1,2,4,5,10; \; \; 30 \rightarrow 1,2,3,5,6,10,15; ...$ but the key ones are $26 \rightarrow 1,2,13$, $105 \rightarrow 1,3,5,7,15$ and $56 \rightarrow 1,2,4,7,8,16$.
(iii) Timeline "death by missing primes": If a link ever reaches a prime number not in any tokens, it is never usable again. For example, in the second puzzle, there is a node with $12$. The max timeline of it is $12 \rightarrow 13 \rightarrow 14 \rightarrow 15 \rightarrow 16 \rightarrow 17 \rightarrow 18 \rightarrow 19$. But as there are no numbers with factor of $17$, the timeline 'dies' at $17$, and the modified timeline becomes $12 \rightarrow 13 \rightarrow 14 \rightarrow 15 \rightarrow 16 $, and it can be passed over max $4$ times.
Solutions
First puzzle: By following the heuristics, the order is unique $18, 8,15,12$. Now, There is only three choice for each: Go through Right/Central/Left path. I brute forced the $3^4$ possibilities in head(most dies off instantly, can be done under 5 minutes), so the $9$ possible solutions are $LRCL, LRCC, LRCR, RLCL, RLCC, RLCR, RRCL, RRRL, RRRR$
Second puzzle: It begins to get annoying from here. I found $17$ solutions. Firstly, after doing the prime factorization, divide into two subcases: Using 10
node, or without using it. If you don't use it, then there's only one entry: $1$, and the timeline of it is $1 \rightarrow 2 \rightarrow3\rightarrow4\rightarrow5\rightarrow6\rightarrow7\rightarrow8$. After following some obvious deduction, the order how the number goes are $1 \rightarrow One \; \; from \; \; the \; \; set \{12,20,30,60\}, 2 \rightarrow 26, 3 \rightarrow 9, 4,5,6 \rightarrow One \; \; from \; \; the \; \; set \{12,20,30,60\}, 7 \rightarrow 105, 8 \rightarrow 56$. Now, there are $11$ ways to take them, and all of them produces a solution !. Doing analogous for the case of using $10$, you first see you can use it only once because there are no numbers with a divisor of $11$. Out of all the possibilites in this case, I checked $6$ to be working, giving a total of $6+11 =17$ possible solutions.
Third puzzle: Analogous to two. I was lazy and annoyed with the lack of near primes, so I didn't check all, but there are $>12$ solutions.
Fourth puzzle: There are many more than 6 solutions. Compute the timeline-image first. The bridge can be passed over only two times, because of no multiple of $7$, the line with $10$ can passed only once, and the bottom right road to exit can be passed on max three times, because of the same reason. Now subcases: Using only $1$ node entry, and using both $3$ and $1$ node entry. Using only $3$ node entry is impossible, because it can be used only $4$ times, and there are $5$ tokens. There are $2^5-2=30$ possibilities for only the entry for this, so I am very lazy for that. For the sub-case Using only $1$ node entry, see the timeline-image (Now you have no choice till the orange dot):

For the order of the tokens, The number has to simultaneously be divided by $1 \rightarrow lcm(1,2), 2 \rightarrow lcm(2,3), 3 \rightarrow lcm(3,4), 4 \rightarrow lcm(4,5), 5 \rightarrow lcm(5,6)$, which gives after some deduction, two possibilities $100/20 , 6, 12, 100/20, 30$. Now notice two must go over the bridge to the left, since the right road to exit can only be passed over $3$ times. There are $4$ possibilities of who goes to the left of the bridge $100/20, 6/12$. Each possibility works, with some sub-cases (of $12$ to rest or not rest on the left side rest), giving a total of $6$ possibilities.
0
. Intentional or not? I ask because division by zero is no good as far as I know. $\endgroup$