Step 1: Recall how the Mohr-Coulomb failure envelope is built. On a shear stress versus normal stress plot, we draw the Mohr circle for the stress state at a point, and the failure envelope is the sloped line \(\tau = c + \sigma \tan\phi\), where \(\phi\) is the angle of internal friction of the soil.
Step 2: Failure happens where the Mohr circle just touches this sloped envelope, not at the very top of the circle. The very top of the circle is where shear stress is largest, equal to the circle's radius, \(\dfrac{\sigma_1-\sigma_3}{2}\).
Step 3: Because the envelope is a straight line tilted upward by the friction angle \(\phi\), the point where the circle touches this tilted line is not the topmost point of the circle unless the line is horizontal, that is, unless \(\phi = 0\). So for any soil with real friction, the shear stress at the point of tangency, the actual failure shear stress, is smaller than the maximum shear stress of the circle. This means Statement (I) does not hold in general.
Step 4: The angle between the failure plane and the major principal plane also depends on \(\phi\), and standard Mohr circle geometry gives this angle as \(45^\circ + \dfrac{\phi}{2}\), not a flat \(45^\circ\). Only when \(\phi = 0\) does this reduce to exactly \(45^\circ\). Since the question does not restrict itself to purely cohesive, frictionless soil, Statement (II) as a general claim also does not hold.
Step 5: Both statements only hold in the special case \(\phi = 0\) and fail to hold in general, so as stated, both Statement (I) and Statement (II) are false.