I would not use Fisher's exact test for your situation. Fisher's exact test assumes the marginals were fixed in advance (see here), which I doubt is true in your case.
I wonder if you are worried about the validity of the chi-squared test. The rule of thumb for the chi-squared test is that the expected values for the cells should be at least 5, not that the observed cell counts need to be. That rule of thumb is known to be too conservative. In your case, four out of six expected values are less than 5, but not by much, so the chi-squared test is probably fine. Nonetheless, since that is a large proportion of your cells, and you have relatively few data, you could set simulate.p.value=TRUE just to be safe:
table(site, infection)
# infection
# site 0 1
# A 1 7
# B 10 2
# C 2 6
chisq.test(table(site, infection))$expected
# infection
# site 0 1
# A 3.714286 4.285714
# B 5.571429 6.428571
# C 3.714286 4.285714
chisq.test(table(site, infection), simulate.p.value=TRUE)
# Pearson's Chi-squared test with simulated p-value
# (based on 2000 replicates)
#
# data: table(site, infection)
# X-squared = 11.75, df = NA, p-value = 0.003498
Another way is to fit a logistic regression model. You can determine if the sites differ by using a likelihood ratio test of the model as a whole (see here). Here we see that the sites differ:
infection = factor(infection, levels=c("0","1"))
mod = glm(infection~site, family=binomial)
summary(mod)
# ...
# Coefficients:
# Estimate Std. Error z value Pr(>|z|)
# (Intercept) 1.9459 1.0690 1.820 0.06872 .
# siteB -3.5553 1.3202 -2.693 0.00708 **
# siteC -0.8473 1.3452 -0.630 0.52878
# ...
# Null deviance: 38.673 on 27 degrees of freedom
# Residual deviance: 25.839 on 25 degrees of freedom
# ...
1-pchisq(38.673-25.839, df=2) # [1] 0.00163355
To run multiple comparisons on the different sites versus each other, we may want to use the Bonferroni correction to the p-value. This is a bit conservative, but may be reasonable since your contrasts are not a-priori or orthogonal. To control the familywise type I error rate at .05, you want the individual p-values to be less than .0167. With relatively few data, you may not want to trust the normal approximation for the Wald tests in the output above. To check for pairwise differences between the sites, you can select only two sites at a time and use the likelihood ratio test as before. From this, you can see that B differs from A, and from C, but that C does not differ from A:
.05/3 # [1] 0.01666667
summary(glm(infection~site, family=binomial, subset=site!="C")) # B vs A
# ...
# Null deviance: 27.526 on 19 degrees of freedom
# Residual deviance: 16.842 on 18 degrees of freedom
# ...
1-pchisq(27.526-16.842, df=1) # [1] 0.001080661
summary(glm(infection~site, family=binomial, subset=site!="A")) # B vs C
# ...
# Null deviance: 26.920 on 19 degrees of freedom
# Residual deviance: 19.811 on 18 degrees of freedom
# ...
1-pchisq(26.920-19.811, df=1) # [1] 0.007669788
summary(glm(infection~site, family=binomial, subset=site!="B")) # C vs A
# ...
# Null deviance: 15.442 on 15 degrees of freedom
# Residual deviance: 15.026 on 14 degrees of freedom
# ...
1-pchisq(15.442-15.026, df=1) # [1] 0.5189397
This is not really different from the standard case of multiple groups with a continuous variable. If you have an a-priori hypothesis that $B < E$, and don't care about the relationships amongst the other groupings, you can simply run a t-test on those two. In your case, the response variable is binary, so you could simply run a $2\times2$ chi-squared test. If you didn't have such an a-priori hypothesis, you don't want to look at the data and notice that $B$ and $E$ are the furthest apart and just test those. In general, it is best to see if there are any differences amongst the groups with a single (ANOVA or chi-squared) test first. That approach provides some protection against type I error inflation. If you don't have an a-priori hypothesis and you want to skip the omnibus test, you definitely need some strategy to hold familywise type I error rates at an acceptable level (e.g., Bonferroni would be one possibility).
Given more information about your situation, what you want is some form of Dunnett's test. The traditional version of Dunnett's test is for a continuous response variable. In your case, you need to do this with a logistic regression model. This pdf provides information on how to do that. Note that it is written for SAS, but you may be able to adapt the procedures for your chosen software.
Best Answer
A contingency table should contain all the mutually exclusive categories on both axes. Inshore/Midchannel/Offshore look fine, however unless "less than 100% mortality" means "100% survival" in this biological setting you may need to construct tables that account for all the cases observed or explain why you restrict your analysis to the extreme ends of the sample.
As 100% survival means 0% mortality, you could have a table with columns 100%=mortality / 100%>mortality>0% / mortality=0%. In this case you wouldn't any more compare percentages, but compare ordinal mortality measures across three site type categories. (What about using the original percentage values instead of categories?) A version of Kruskal-Wallis test may be appropriate here that takes ties appropriately into consideration (maybe a permutation test).
There are established post hoc tests for the Kruskal-Wallis test: 1, 2, 3. (A resampling approach may help tackling with ties.)
Logistic regression and binomial regression may be even better as they not only give you p values, but also useful estimates and confidence intervals of the effect sizes. However to set up those models more details would be needed concerning the 100%>mortality>0% sites.