I have no doubt that an appropriate regression model might be more elegant and might allow for testing more details than can be done with a couple of t tests. However, it seems to me that the two most important issues are (a) whether the treatment group improved from the first exam to the second, possibly on account of learning something in the course. and (b) whether the treatment group improved more than the control group from the first to the second exam.

If you have about 30 students in each group and the exams have about 50 T-F questions each, then the scores might be nearly normal so that (a) could be answered with a paired t test (or a one-sample t test on differences, and (b) could be answered with a two-sample t test. With such sample sizes and no indications of serious departures from normality, my first analyses would be to do these two tests to get an idea whether the major expectations of the study are met.

I have simulated some fake data for purposes of illustration. Let the improvement scores for treatment and control groups be td and cd, respectively.

summary(td)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
   5.00    7.25   10.00    9.40   11.00   14.00 
summary(cd)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
  0.000   3.000   4.000   4.367   5.750   9.000 

For my fake data, boxplots (Treatment=1) show no serious skewness or outliers.

boxplot(td, cd, horizontal=T)

Normal probability plots show discreteness but no serious departures for normality.

par(mfrow=c(1,2))
 qqnorm(td,main="Treatment: Normal Probability Plot"); qqline(td)
 qqnorm(cd,main="Control: Normal Probability Plot"); qqline(cd)
par(mfrow=c(1,1))

Thus, it seems reasonable to compare 'improvement' scores with a two-sample t test. The P-value is very nearly $0$ and we reject the null hypothesis (of no difference).

t.test(td, cd)

       Welch Two Sample t-test

data:  td and cd
t = 8.5125, df = 57.587, p-value = 9.006e-12
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
3.849569 6.217098
sample estimates:
mean of x mean of y 
9.400000  4.366667

Also, a right-sided, one-sample t test on the improvement scores in the treatment group, finds statistically significant improvement.

t.test(td, alt="gr")

        One Sample t-test

data:  td
t = 23.5, df = 29, p-value < 2.2e-16
alternative hypothesis: true mean is greater than 0
95 percent confidence interval:
 8.720349      Inf
sample estimates:
mean of x 
      9.4 

Addendum: In your Question, it seems that you might not have exam scores; just whether subjects passed or failed the exams. What's above was written hoping you do have scores. But even if not, I suppose you know, whether a subject improved (went from fail to pass), stayed the same, or went from pass to fail.

Then for Treatment subjects you might have 3 down, 5 same, and 22 up. You could do a proportion test to see if the proportion who improved is significantly better than half. With these counts the answer is Yes. [These hypothetical counts are not derived from the simulated data used for the t tests.]

prop.test(c(22, 8), c(30,30), alt="gr", cor=F)

        2-sample test for equality of proportions 
        without continuity correction

data:  c(22, 8) out of c(30, 30)
X-squared = 13.067, df = 1, p-value = 0.0001503
alternative hypothesis: greater
95 percent confidence interval:
 0.2788575 1.0000000
sample estimates:
   prop 1    prop 2 
0.7333333 0.2666667

By contrast, if the Control group had 4 down, 21 same, 5, improved, then you could do a chi-squared test to see if the Treatment group did significantly better. The result is that Treatment and Control groups are clearly not homogeneous. (The simulation option was used because the cells for subjects going from pass to fail have so few counts.)

MAT = cbind(c(3, 5, 22), c(4, 21, 5))
chisq.test(MAT, sim=T)

        Pearson's Chi-squared test with simulated p-value 
        (based on 2000 replicates)

data:  MAT
X-squared = 20.693, df = NA, p-value = 0.0004998

Fisher's exact test is another option, also with a highly significant result.

fisher.test(MAT)

        Fisher's Exact Test for Count Data

data:  MAT
p-value = 1.123e-05
alternative hypothesis: two.sided

Note: The data for t tests were sampled in R according to the program below.

set.seed(301)
t1 = rbinom(30, 50, .5)
td = rbinom(30, 50, .2)
t2 = t1 + td
c1 = rbinom(30, 50, .5)
cd = rbinom(30, 50, .1)-1
c2 = c1 + cd