Given ABC = 50^ and ACB = 60^.

The sum of angles in a triangle is $180^\circ$.
Given $\angle ABC = 50^\circ$ and $\angle ACB = 60^\circ$.

Since A is due north of C, the line segment CA is a North-South line.
Draw a North line at B, parallel to CA. Let $N_B$ be the North direction from B.
The angle $\angle ACB = 60^\circ$ and the angle between the South line from B (let's call it $S_B$) and BC are alternate interior angles.
So, the angle from the South line at B to BC is $60^\circ$.
The bearing of C from B is measured clockwise from the North line at B. This is $180^\circ$ (to the South line) plus the $60^\circ$ angle.

The bearing of A from B is the bearing of C from B minus the angle $\angle ABC$, because A is $50^\circ$ counter-clockwise from C, relative to B.

The bearing of A from B is $\boxed{190^\circ}$.

Since A is due north of C, the line segment AC is a North-South line. The North direction from A points away from C. The line segment AC itself represents the South direction from A.

The bearing of B from A is measured clockwise from the North line at A to the line AB.
The angle from the North line at A, clockwise to the South line at A (which is AC), is $180^\circ$.
We found $\angle BAC = 70^\circ$. This angle is between the South line (AC) and the line AB.
Since AB is $70^\circ$ to the "left" (west) of the South line (AC), we subtract this angle from $180^\circ$.

The bearing of B from A is $\boxed{110^\circ}$.