I think I have managed to determine that the answer to number 1 is C...
Note we have a 3-way branch at A, and 2-way branchings at C and D, all other nodes have no choice of path.
So our possible paths start:
AB
AC
AD
the first path has only one completion:
ABEH
the next path can go one of two ways:
ACE...
ACF...
after that, we can only complete each of THOSE one way:
ACEH
ACFH
finally, we can go one of two ways after AD:
ADF...
ADG...
and these both can be completed only one way:
ADFH
ADGH, giving 5 paths in all:
ABEH
ACEH
ACFH
ADFH
ADGH
To compute the time required for completing the task, we add the times required for each individual task:
Time for ABEH, for example, is 3+7+8+5 = 23 days.
Calculate the times for the other 4 complete task paths, the critical path is the one with the shortest completion time.
To find the critical path for F, treat F as if it were the "end-stage" of the assembly product.