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The null hypothesis states that there is no difference between cars with experimental fuel injection system cars with the current fuel injection system.
The alternative hypothesis states that cars with experimental fuel injection system had better mileage.

Sample 1 size 12
Sample 2 size 12
Sample 1 mean 33.4
Sample 2 mean 28.6
Sample 1 S.D. 1.3
Sample 2 S.D. 2
Degree of freedom =
[s1^2/n1+s2^2/n2]^2 / [(s1^2/n1)^2/(n1-1)+(s2^2/n2)^2/(n2-1)]
= [1.69/12 + 4/12]^2 / [(1.69/12)^2 /11+ (4/12)^2/11]
Denominator of t = 0.688598
t = (33.4 – 28.6) = 4.8 / 0.688598
t =6.9707
Degree of freedom 18.8871

Compare the t-value of 6.97 with critical t (with 19 degrees of freedom, 5% level of significance)
t-critical value = 1.729

Since computed t exceeds critical t, reject the null hypothesis.

b)
Conclusion:
We rejected the null hypothesis
cars with the experimental fuel injection system provide better gas mileage than cars with the current fuel injection system.

c)
Let d be the difference in gas mileage between the two systems.
H0: d=0
HA: d > 0
We would use a paired one sample t-test (dependent t-test) and consider the difference in mileage as our sample.

Note:
I am confused as to your wording ‘single-blind.’
Double blind means neither the experimenter nor the subjects( people who choose the cars) know which unit is selected and assigned to which group. In a single blind study, the person who divides them into two groups knows which group each car is assigned to.