Let's analyze the given regular expression: $0(0+1)^*(0+1)+ϵ+1(0+1)^*1$
Breaking it into 3 parts:
$0(0+1)^*(0+1)\ $: This consists of strings that start with a 0 and can end with either a 0 or a 1. In between we can have any arbitrary length of string in {0, 1} (i.e $(0+1)^*$). The minimum length strings are 01 and 00.
$\epsilon \ $: This is the empty string. Hence the language contains empty string.
$1(0+1)^*1\ $: This represents the set of strings starting with a 1 and ending with a 1, with anything in between. Here the minimum length string is 11.
So, the language contains strings from all these parts.
"All the strings including empty string of length >=2, if started with 1 should end with 1"
= All strings of length greater than or equal to 2 which start with 1 and end with a 1 .
+ All strings of length greater or equal to 2 starting with a 0.
+ epsilon (empty string).
