Change in momentum?

Momentum is typically in units of kg*m/s. Therefore the 80 km/h should be converted to m/s before going on to momentum.
#80 (km)/h * ((1000 m)/(1 km)) * ((1 h)/(3600 s)) = 2.22 m/s#

Yes, p=mv. The initial momentum is
#p_i = 1500 kg * 2.222 m/s = 3333 kg*m/s#

Since the final velocity is zero,
#p_f = 0#

The change is
#Deltap = p_f - p_f = 0 - 3333 kg*m/s = -3333 kg*m/s#
Because of the rules regarding the number of significant digits, the answer to this part should be given as -3000 kg*m/s.

In the case of the accident , the initial momentum is
#p_i = 1500 kg * 2.222 m/s = 3333 kg*m/s#

Since the final velocity is zero,
#p_f = 0#

The change is #p_f - p_f = 0 - 3333 kg*m/s = -3333 kg*m/s#
Because of the rules regarding the number of significant digits, the answer to this part should be given as -3000 kg*m/s.

The times that were given in the question were not relevant to what the question asked for. However, those times would be used if it also asked for what the average force was during those 2 cases.

Additional analysis: The average force during those 2 cases:

Change in momentum calculated above is called impulse. Another formula for impulse is calculated by multiplying average force by the time of application of that force. Both methods yield impulse, so we can write
#Deltap = F_"ave"*t#

We know that #Deltap = -3000 kg*m/s#, so we can solve for #F_"ave"#.

The case of breaking to a stop in 4 s:

#Deltap = -3000 kg*m/s = F_"ave"*4 s#

#F_"ave" = (-3000 kg*m/s)/(4 s) = -750 kg*m/s^2 = -750 N#

The case of an accident stopping the car in 0.25 s:

#Deltap = -3000 kg*m/s = F_"ave"*0.25 s#

#F_"ave" = (-3000 kg*m/s)/(0.25 s) = -12,000 kg*m/s^2 = -12,000 N#

I hope this helps,
Steve