Bit mask in Bash

Yes, this is entirely possible:

#!/bin/bash

var1=0xA # (0b1010)
if (( (var1 & 0x3) == 0x2 ))
then
  echo "Match"
fi

Bit manipulation is supported in POSIX arithmetic expressions:

if [ $(( var1 & 0x3 )) -eq $(( 0x2 )) ]; then

However, it's a bit simpler use an arithmetic statement in bash:

if (( (var1 & 0x3) == 0x2 )); then