1) Initial upward acceleration: 
2) Mass of burned fuel: 
Explanation:
1)
There are two forces acting on the rocket at the beginning:
- The force of gravity, of magnitude
, in the downward direction, where
is the rocket's mass
is the acceleration of gravity
- The thrust of the motor, T, in the upward direction, of magnitude

According to Newton's second law of motion, the net force on the rocket must be equal to the product between its mass and its acceleration, so we can write:
(1)
where a is the acceleration of the rocket.
Solving for a, we find the initial acceleration:

2)
When the rocket reaches an altitude of 5000 m, its acceleration has increased to

The reason for this increase is that the mass of the rocket has decreased, because the rocket has burned some fuel.
We can therefore rewrite eq.(1) as

where
is the new mass of the rocket
Re-arranging the equation and solving for m', we find

And since the initial mass of the rocket was

This means that the mass of fuel burned is
