4: In total, the taxi moves 7 blocks north and 5 east. Ask Pythagoras for the diagonal of the rectangle.

5,6: Use components. The x-comp of A is +3, the y-comp is 0. The x-comp of B is -2sqrt(3) (=-4cos30), the y-comp is +2 (=4sin30). Add them. The resultant vector has an x-comp = 3-2sqrt(3), and y-comp = +2. Use Pythagoras to get the magnitude, and x/y=tan(a) to get the angle. You may have to add 90deg; make a little sketch:

7,8: These two problems use the independance of horizontal and vertical movements. There is no horizontal acceleration in free flight, and the vertical acceleration is always g = -9.8 m/s².

The horizontal velocity component is 29.4cos30, so the distance traveled is 3*29.4cos30.
The time to the maximum height is t =1.5 s, and the distance the ball falls in the last half of the flight is 0.5*gt².

10: The initial vertical velocity is zero.
The time interval is the free-fall time to drop 10 cm =0.1 m (find using y =0.1 =0.5gt²).
The required velocity is then the horizontal distance / time interval.

11. This is a "sum of vectors" problem. Set up the coordinate system E-W, N-S. The components of the plane vector in still air are v_pE =500 mph, v_pN =0 mph. The components of the wind vector which is carrying the plane are v_wE =104 mph (120*cos30), v_wN =60 mph (120*sin30). Add these components to get the resultant components: v_p+wE =604, v_p+wN =60. From these components one can recover the magnitude (Pythagoras) and the angle (sin or tangent).

12. This is also a vector addition problem, but now the vectors are all in a line so we don't have to worry about the "cross-river" direction.
The boat moves downstream at 10+2 =12 m/s, and upstream at 10-2 =8 m/s relative to the land. So the time to go 1000 m downstream is 1000/12 s, and the time to go 1000 m upstream is 1000/8 s.

13. It is always a good idea to draw a diagram to fix the question and the "knowns".

Also remember the horizontal motion here has no acceleration, so an appropriate formula will be v_x=Dx/Dt. We will use this formula to determine the time interval between bat and fence: Dt =132/v_x =132/(44*cos30). The question asks about a height at the point in the path above the fence. Heights can best be found with the formula y =y₀+v_y0t+0.5gt². Here y₀ =1 m, and v_y0 =22 m/s (44*sin30). Using the time interval from the previous step, we can calculate the height at the fence, =18.4 m, 13.4 m above the top of the fence.