given y[n]=x[2n]

the value of y[n] does not depend upon the values of x[k] for k<n hence it is memoryless

x[n]=y[n/2] since x[n] can be expressed in terms of y[n] it is INVERTIBLE

for a function x1[n] the output is y1[n]=x1[2n]

for a function x2[n] the output is y2[n]=x2[2n]

consider a function x3[n]=x1[n]+x2[n]

the output is y3[n]=x3[2n] = x1[2n]+x2[2n] = y1[n]+y2[n]

since for a function x1[n]+x2[n] the output is y1[n]+y2[n] it is LINEAR

for a function x[n] the output is y[n]=x[2n]

consider function x1[n]=x[n-t] the output y1[n]=x[2(n-t)]

if u delay the function y[n] by t u get x[2n-t] since they both are not equal it is NOT TIME INVARIANT

since the value of y[n] depends upon the future values of x[n] it is NOT CASUAL

if we give a bounded input x[n] for all values of n then x[2n] is also bounded so is y[n] hence it is BIBO STABLE

Is this corrct for question a?