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Cool, you just changed the output record separator accordingly.
I am just using the tools, while you are playing with the tools.
That was amazing!

awk 'BEGIN{OFS="\t"}
NR == FNR{a[$0]; next}
FNR == 1{ for(i=1;i<=NF;++i){if($i in a){a[$i]=i}} for(e in a){printf e OFS} printf ORS; next}
{for(i in a){printf $a[i] OFS} printf ORS}'...

awk 'NR>=3 && /red/' infile

perl:

You probably need unix2dos and dos2unix.
Windows uses 0x0d0a as newline, while Unix uses 0x0a.

If you want to list all files matching the pattern under /etl/PDAC/CBA/Supplier/, use the following script:
#!/usr/bin/perl
use strict;
use warnings;
opendir(DIR, "/etl/PDAC/CBA/Supplier/") or...

You can read multiple fields:
read month file;
May filename.txt<ENTER>;
grep "$month" "$file"

OK, I guess you still have more than 2 identical consecutive words spanning multiple lines, and you have to reserve all the newline characters, this can be a challenge, at least with SED.
...

Fixed, I forgot to add the word boundary(marked as red) for the identical consecutive word:

sed ':f;N;$!bf; s/\b\(.*\)\n\1\b/\1\n/g; s/\b\(.*\)\b\1\b/\1/g'

You'll have to use backticks:
./script.sh > `date +"%Y%m%d"`

echo "DATA" | grep "^[DH]"

pbillast's solution won't work as expected when Name or Country contains more than one word, the following code does:
awk '/^(Name|Age|Country)/{ $1=""; sub(/^[ \t]+/, "");...

You may stick to grep and cut:
echo "/home/one/two/three/" | grep -Eo '[^/]+/?$' | cut -d / -f1
Or just perl:
echo "/home/one/two/three/" | perl -pe 's|^.+?([^/]+)/?$|$1|'

If AWK is an option:
gawk 'FNR==NR{a[$0]; next} {if($0 in a){delete a[$0]}else{++c}} END { print "Count of difference: ", length(a)+c }' data.txt data2.txt

Good idea, but this will be better.
grep ", *," infile

awk -F, ' {for(i=1;i<=NF;++i) { s=$i; gsub(/ +/, "", s); if(s=="") { print; break; }} } ' infile OFS=,

try this:

echo "Youcaneasilydothisbyhighlightingyourcode." | sed 's/\(.\{3\}\)/\1 /g'

awk -F: '{print $1}' infile

If you remove the ".*" part from the regex, that means you want to match a line that has only 16 characters, this simply does not match anything, so the original string is output.
...

There are punctuations in your string, so:
awk -F"[ ,.]" '{for(i=1;i<=NF;++i) if($i!="")a[$i]} END {for(s in a) print s}' infile
If you insist on using FS, try this, it has the same output as the...

Or this:
awk -F- '{print $1}' infile

try this:
awk ' { ++a[$0] } END { for(s in a) { for(i=1;i<=a[s];++i) { print s, i, a[s] } } } ' inputfile

Note: your input data is assumed to be sorted.

The AWK way:
awk '/^[0-9]+/ { print "63"$0; next } { print }' infile