There are no appropriate units for power on the list you provided
4. The Coyote has an initial position vector of
.
4a. The Coyote has an initial velocity vector of
. His position at time
is given by the vector

where
is the Coyote's acceleration vector at time
. He experiences acceleration only in the downward direction because of gravity, and in particular
where
. Splitting up the position vector into components, we have
with


The Coyote hits the ground when
:

4b. Here we evaluate
at the time found in (4a).

5. The shell has initial position vector
, and we're told that after some time the bullet (now separated from the shell) has a position of
.
5a. The vertical component of the shell's position vector is

We find the shell hits the ground at

5b. The horizontal component of the bullet's position vector is

where
is the muzzle velocity of the bullet. It traveled 3500 m in the time it took the shell to fall to the ground, so we can solve for
:

Molarity and molality both describe the concentration of a substance in terms of moles.
Molarity describes the number of moles of a substance per unit of volume, typically per liter (mol/l).
Molality describes the number of moles per unit of mass, typically kilograms (mol/kg).
When determining the molality of a solution, mol/kg can be obtained by finding the number of moles in the substance, and dividing that number by the the total weight in kilograms of that substance.
When determining the molarity of a solution, mol/l can be obtained by dividing the number of moles in a substance by the total volume in liters of that substance.
1) 3 miles/Hour
The speed is defined as the distance covered divided by the time taken:

where
d = 1.5 mi is the distance
t = 0.5 h is the time taken
Substituting,

2) 1.34 m/s south
Velocity, instead, is a vector, so it has both a magnitude and a direction. We have:
is the displacement in meters
is the time taken in seconds
Substituting,

And the direction of the velocity is the same as the displacement, so it is south.