removing lines with special characters in awk

how can I fix it?

You did

awk -F "\t" '{ if(($1 == '*')) { print $1 "," $2} }' infile.txt > outfile.txt

by doing $1 == "*" you are asking: is first field * not does first contain *? You might use index function which does return position of match if found or 0 otherwise. Let infile.txt content be

VAREAKAVVLRDRKSTRLN 2888
ACP*VRWPIYTACGP 292
RDRKSTRLNSSHVVTSRMP 114
VAREA*KAVVLRDRRAHV*T    73

then

awk 'index($1,"*")==0{print $1,$2}' infile.txt

output

VAREAKAVVLRDRKSTRLN 2888
RDRKSTRLNSSHVVTSRMP 114

Note that if you use index rather than pattern /.../ you do not have to care about characters with special meaning, e.g. .. Note that for data you have you do not have to set field separator (FS) explicitly. Important ' is not legal string delimiter in GNU AWK, you should use " for that purpose, unless your intent is to summon hard to find bugs.

(tested in gawk 4.2.1)

with your shown samples, please try following awk program.

awk '$1!~/\*/' Input_file

OR above will print complete line when condition is NOT matched, in case you want to print only 1st and 2nd fields of matched condition line then try following:

awk '$1!~/\*/{print $1,$2}' Input_file

Use grep like so to remove the lines that contain literal asterisk (*). Note that it should be escaped with a backslash (\*) or put in a character class ([*]) to prevent grep from interpreting * as a modifier meaning 0 or more characters:

echo "A*B\nCD" | grep -v '[*]'
CD

Here, GNU grep uses the following options:
-v : Print lines that do not match.