Proportion of Intervals from t-interval method Here is an excerpt of a book question:

The t-interval method assumes that the underlying distribution is
  approximately normal. If we were to generate 800 samples of sample
  size 15 from this exponential distribution and the calculate 95%
  t-intervals, what is the proportion of intervals that include the true
  mean 900? Is this what you would expect? Explain.

Here is the R code for 800 samples.
sims = 800
n = 15
tsamples <- replicate(sims, rexp(n, rate = lambda))
mu = 1/lambda

results <- as.numeric(sims)
t.int <- matrix(FALSE, sims, 2)
for (i in 1:sims) {
    t.int[i, ] <- t.test(tsamples[, i], conf.level = 0.95)$conf.int
    results[i] <- t.int[i, 1] < mu & t.int[i, 2] > mu
}
sum(results)/sims

Output changes every time but a few are 0.9175, .91, .905, .925
My question is, I don't truly understand what it means by asking for "proportion of intervals that include the true mean 900". How can I figure out what this is asking and what the answer is? 
Is it asking how many times the R result is within a certain amount? Am I supposed to run the code multiple times with sims = 900 and find the lowest and highest number for a range?
 A: Every time you generate an interval, it's a Bernoulli trial. Either the interval includes the true mean or it doesn't; the trials are independent and the (a priori) probability is constant.
So when you generate 800 such intervals, you conduct 800 Bernoulli trials with an unknown population proportion, $p$ (being the actual coverage probability of a randomly selected interval).
Could you estimate $p$ from the data? Can you give an interval for $p$ from the data?
A: By definition an (approximate) P% confidence interval is a random interval that includes the mean  (approximately) P% of the time. This should guide what you should expect from a 95% t-interval method that assumes that the underlying distribution is (approximately) normal.

My question is, I don't truly understand what it means by asking for "proportion of intervals that include the true mean 900". How can I figure out what this is asking and what the answer is?

To see what this is asking, I've augmented your code to plot the confidence intervals (see comments in the code).
lambda = 1/900 # initialize lambda so that mean is 900

sims = 100 # reduce sims to make plot comprehensible
n = 15
tsamples <- replicate(sims, rexp(n, rate = lambda))
mu = 1/lambda

results <- as.numeric(sims)
t.int <- matrix(FALSE, sims, 2)
ptsx = c() # empty list of confidence intervals
for (i in 1:sims) {
    t.int[i, ] <- t.test(tsamples[, i], conf.level = 0.95)$conf.int
    results[i] <- t.int[i, 1] < mu & t.int[i, 2] > mu
    ptsx = c(ptsx, t.int[i, 1], t.int[i, 2], NA) # add to list
}
sum(results)/sims

# plot confidence intervals
plot(0, xlab="x", ylab="i", xlim=c(min(0,t.int[, 1]), max(t.int[, 2])), ylim=c(0,sims))
points(ptsx, rep(1:sims,each=3), type="l")
points(ptsx[rep(results,each=3)==0], rep((1:sims)[results==0],each=3), type="l",col="red", lwd=2)
abline(v=mu,col="green")

The output from a typical run looks like the image below. In this case sum(results)/sims printed out 0.92.

In this plot, the true mean is indicated by the vertical green line. Confidence intervals that don't contain the mean are plotted in red: all confidence intervals are plotted in black, using the first points command, and then the red confidence intervals are overplotted using the fact that the results vector contains 1s where confidence intervals contain the mean and 0s otherwise.
You get the proportion of intervals by counting the number of intervals that contain the mean (on the plot it's easier to count the ones that don't and subtract that from 100) and dividing by the total number of intervals. The code already calculates it for you.
As to what the question's really asking, which is the "Explain bit", the plot provides a clue in that all the red intervals are to one side of the mean. To investigate further, I would plot histograms (or similar) of the columns of data in tsamples that led to the confidence intervals that don't contain the mean. Your first port of call when you don't understand something should be to plot it if you can.
