You are right that one way to solve this problem is with the principle of inclusion and exclusion.
For the "at least 3 consecutive digits" case, note that there are 8 possible sets of 3 consecutive digits. These digits can be at the right of the pin (like X234) or at the left (234X), where X can be any digit. This gives $8 \cdot 2 \cdot 10 = 160$ pins. However, we have now counted pins with 4 consecutive digits 2 times: for example, 1234 is counted as 123X and X234. Thus we subtract the number of these pins which is 7 to get $153$ pins with at least 3 consecutive digits.
You can extend this method to the second part too. Add all pins with 2 known consecutive digits, subtract some with 3 known consecutive digits, add some with 4 consecutive digits, where "some" is chosen such that each of those categories is counted exactly once.