i think this problem is analgous to a person being accelerated in an elevator so it might help if you think about it like that.

it says that the pilot is pulling out of a dive, meaning that he is no longer "free-falling" but pointing (or trying to) the nose of the plane against gravity. you can imagine this in an evelator that is constantly accerlating you against gravity.

if you do a free body dia. and do fnet=ma you'll see that the normal force (thus the apparent weight) will increase when going against gravity. so if you assume this evelator is going up 3 m/s^2 then you'll have the equation: N-mg=ma, where N is the normal force that youre trying to solve for. you get N= m(10+3) so the normal force is higher than if just free-falling or "diving".

in the same case if the pilot decided to accelerate while diving, or the elevator was accelerated downward with a force outside gravity, the apparent weight would decrease. N-mg=m(-a) -> N= m(g-a) so you would weigh less if you were in this elevator. hope that helps!