#### Answer

The heading of $Z$ from $X$ is $~~75^{\circ}$
The distance $YZ$ is $~~273.6~miles$

#### Work Step by Step

The points $X$, $Y,$ and $Z$ form a triangle.
We can find $\angle XYZ$:
$\angle XYZ = 35^{\circ}+(180^{\circ}-145^{\circ}) = 70^{\circ}$
Since $XY = XZ = 400~mi$, the triangle is an isosceles triangle.
Then $\angle XZY = \angle XYZ = 70^{\circ}$
We can find $\angle YXZ$:
$\angle YXZ = 180^{\circ} - \angle XYZ - \angle XZY$
$\angle YXZ = 180^{\circ} - 70^{\circ} - 70^{\circ}$
$\angle YXZ = 40^{\circ}$
We can find the heading of $Z$ from $X$:
$35^{\circ}+40^{\circ} = 75^{\circ}$
The heading of $Z$ from $X$ is $~~75^{\circ}$
We can find the distance $YZ$:
$\frac{YZ}{sin~\angle YXZ} = \frac{XY}{sin~\angle XZY}$
$YZ = \frac{XY~sin~\angle YXZ}{sin~\angle XZY}$
$YZ = \frac{400~sin~40^{\circ}}{sin~70^{\circ}}$
$YZ = 273.6~mi$
The distance $YZ$ is $~~273.6~miles$