motion....

3)
vector equation for the projectile of B
is
S_B=(10t√3)i + (10t-5t²+10) j

S_A=(10√3-10√3t) i + (10t-5t²) j

S_A-S_B=((10√3-10√3t-10t√3)-10j

minimum at t=0.75s

Min|S_A-S_B|=10

2)
first boat goes to B after 1 hr 20 mins hence B should have 80/20 boats hence 4 boats
first boat reaches A at2hr20 mins hence A should have 140/20 hence 7 boats
therefore total boats between ports is 4+7=11

14b)?

simpler way for 3)

note that vertical component of velocity for both the particles is same, hence vertical distance between them is always 10m.
so the shortest distance will be when the horizontal distance between them is zero.
.:. answer = 10m