(i) Impulse response of the system:
By z-transform,
Y(z)=34z−1Y(z)−18z−2Y(z)+X(z)Y(z)−34z−1Y(z)+18z−2Y(z)=X(z)Y(z)(1−34z−1+18z−2)=X(z)H(z)=Y(z)X(z)=11−34z−1+18z−2H(z)=z2z2−34z+18H(z)=z2(z−12)(z−14)H(z)z=z(z−12)(z−14)
Solving the above equation using partial fractions
H(z)z=A(z−12)+B(z−14)

∴H(z)z=2(z−12)−1(z−14)∴H(z)=2(1−12z−1)−1(1−14z−1)
By Inverse z-transform,

Step response of the system:
x(n)=u(n) …………………(step response)
X(z)=11−z−1 …………………(By z-transform)
we know, H(z)=Y(z)X(z)∴Y(z)=H(z).X(z)=1(1−34z−1+18z−2∗11−z−1=z2z2−34z+18∗zz−1Y(z)z=z2z−1⁄2)(z−1⁄4)(z−1)
By partial fraction,
Y(z)z=A(z−1⁄2)+B(z−1⁄4)+C(z−1)

Y(z)z=−2(z−1⁄2)+1/3(z−1⁄4)+8/3(z−1)Y(z)=−2z(z−1⁄2)+z/3(z−1⁄4)+8z/3(z−1)
By Inverse z-transform,
y(n)=−2(12)nu(n)+13(14)nu(n)+83(1)nu(n)
(iii) The system function is given by,
H(z)=z2(z−12)(z−14)
Poles at: z=1/2, z=1/4.
Zeros at: z=0, z=0.
Pole zero diagrams:

Since all the poles lie inside the unit circle, the system is stable.