Exercise 1: why doesn't this example print anything?

   1 #!/bin/bash
   2 set -e
   3 i=0
   4 let i++
   5 echo "i is $i"

According to the manual, set -e exits "if a simple command (see SHELL GRAMMAR above) exits with a non-zero status. The shell does not exit if the command that fails is part of the command list immediately following a while or until keyword, part of the test in a if statement, part of an && or || list, or if the command's return value is being inverted via !".

The let command is a simple command, and it doesn't qualify for any of the exceptions in the above list. Moreover, help let tells us "If the last ARG evaluates to 0, let returns 1; 0 is returned otherwise." i++ evaluates to 0, so let i++ returns 1 and trips the set -e. The script aborts. Because we added 1 to a variable.

Exercise 2: why does this one appear to work?

   1 #!/bin/bash
   2 set -e
   3 i=0
   4 ((i++))
   5 echo "i is $i"

((...)) does not qualify as a simple command according to the shell grammar. So it is not eligible to trigger a set -e abort, even though it still returns 1 in this particular instance (because i++ evaluates to 0 while setting i to 1, and because 0 is considered false in a math context).

However, this behavior changed in bash 4.1. Exercise 2 works only in bash 4.0 and earlier! In bash 4.1, ((...)) qualifies for set -e abortion, and this exercise will print nothing, the same as Exercise 1.

This reinforces my point about how unreliable set -e is. You can't even count on it to behave consistently across point-releases of a shell.