1. An air parcel at the surface is non-saturated and has a temperature of 30 deg C and a Tdew of 25 deg C. Assuming it moves adiabatically and the SALR = -8.5 deg C/1 km what will be the air temperature at 3000 m?

Air is initally dry so we need to find when the air has cooled down to Tdew (i.e. Tair=Tdew).
DALR =-10 deg C /1 km

T0=30 deg C z0= 0 m

T1= 25 deg C z1=? m

DALR = T1- T0
       z1-z0

- 10 = 25 - 30
1000   z1 - 0

- 10 = -5
1000   z1

1000= z1
-10   -5

1000 * -5 = z1
-10

500 m = z1

SALR =-8.5deg C /1 km

T1=25 deg C z1= 0 m

T2= ?deg C z2=3000 m

SALR = T2- T1
       z2-z1

-8.5 = T2- 25
1000   3000 - 500

- 8.5= T2- 25
1000   2500

- 8.5 *2500= T2- 25
1000

-21.25 = T2-25
 

-21.25 +25= T2

3.75 deg C = T2

Temperature at 3000 m is 3.75 deg C

2. What would be the temperature at 3000 m if the SALR was -6.5 deg C/1000 m? (i.e. assuming the only thing that changed from the above problem is the SALR)

SALR =-6.5deg C /1 km
T1=25 deg C z1= 0 m

T2= ?deg C z2=3000 m

SALR = T2- T1
       z2-z1

-6.5 = T2- 25
1000   3000 - 500

- 6.5= T2- 25
1000   2500

- 6.5 *2500= T2- 25
1000

-16.25 = T2-25
 

-16.25 +25= T2

8.75 deg C = T2

Temperature at 3000 m is 8.75 deg C

3. Why is it so much warmer ?

Because of the additional release of Latent Heat of vaporization when the water condenses.