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For output1:

awk '/;;WORD/ { print $2 }
/ lex:/ {
sub(/.*:[+]/,"")
gsub("/[^+]*[+]", " + ")
sub("/[^+]*$","")
print }' infile

For output 2:

awk '/;;WORD/ {printf "%s%s",...

Another one:
awk '{split($0,F); gsub(/_\+/,x); for(i=2;i<=NF;i++) if($1==$i) $1=$1 OFS F[i]}NF=1' file
(will work in most awks, but not all)

awk '{s=$1; split($0,F); gsub(/_\+/,x);...

OK, good. Here are a couple of ways to do it:
sed '/^[^\/]*$/d;s|/[^+]*+|+|g;s|/.*$||g;s/^+//;s/+$//;s/+/ + /g' input_fileand
awk '
index($0, "/") { # for any line containing a "/"...

Try: perl -nle 'print join " + ", map {s/[\+\/]//g;$_} /\+.*?\//g' file

This will remove blank lines and those which only contain spaces...

grep -v '^ *$' $files

If you wanted to use sed, then...

sed '/^ *$/d' $files