Answer is 1)
1) To accept the input we will push "X" for "0" and when we will encounter "1" sim ply pop it.
In case M=N it will be same as 0^N 1^N or M>N then pop every "1" after "0".
2) here we required two memory comparison 1st for i=2j and j =2k , so this is not CFL.