$$a) H_1 (z) = 6 + z^{-1} - z^{-2}$$

$H_1(z)=\dfrac {6z^2+z-1}{z^2}=\dfrac {6(z+1/2)(z-1/3)}{z^2}\\ zeros \space \space z_1=\dfrac {-1}2,z_2=\dfrac 13$

As both zeros are inside unit circle, this system is minimum phase system.

$b) 1 -z^ {-1} -6z^{-2} \\ H_2(z)=1-z^{-1}-6z^{-2}\\ =\dfrac {z^2-z-6}{z^2}=\dfrac {(z+2)(z-2)}{z^2}\\ \text {zeros are at } z_1=-2, …