A person facing West moves 30 m towards West, then 80 m North, then 30 m towards West again. How far is the final position from the starting point?

East–West total: 30 + 30 = 60 m west.
North–South total: 80 m north.
Distance = √(60² + 80²) = √10000 = 100 m.
Faster: 60 and 80 is the 3-4-5 triple scaled by 20, so the answer is 5 × 20 = 100 without squaring anything.

A goes 500 m straight from his house, turns left and walks 1 km, then turns right and walks 1½ km to reach his office. If the office lies North-East of the house, in which direction did A start, and what is the shortest distance from house to office?

The finishing position is North-East of the start, so both the net north and net east components must be positive. Test East as the starting direction:
500 m East → turn left, now facing North, 1 km → turn right, now facing East, 1.5 km.
Net East = 0.5 + 1.5 = 2 km; net North = 1 km. Both positive, so North-East checks out.
Distance = √(2² + 1²) = √5 km, starting towards East.
Method: when the starting direction is unknown, use the stated final bearing as your test. Trying East first is usually quickest.

A person starts at A, travels 13 m North-East, then 12 m South, then 13 m North-East again, then 10 m West, reaching B. Find the direction of A from B and the distance between them.

The 13 and 12 are the giveaway — each 13 m north-easterly leg is meant as the 5-12-13 triple, i.e. 5 m east together with 12 m north.
Leg 1: +5 E, +12 N. Leg 2: −12 N (cancels exactly). Leg 3: +5 E, +12 N. Leg 4: −10 E.
Net East = 5 + 5 − 10 = 0. Net North = 12.
So B lies 12 m due north of A, which means A is South of B, 12 m away.
Note: the question asks for the direction of A from B, not B from A. Reversing the two is the trap, and "North, 12 metres" is sitting in the options for anyone who does.