How to find farthest see-able corners of a rectangle? I don't know English well enough to explain so here a picture that explains. This is on a coordinate plane.

 A: A way to solve this problem is to compute the convex hull of the four corners and the viewpoint (see this, for example).  The two vertices adjacent to the viewpoint are the solution.  If the viewpoint is on the extension of one side, you need to define which of the two vertices of that side is the farthest visible.
A: Considers the angles $\theta_i$ defined by all vectors $\vec{OA},\vec{OB},\vec{OC},\vec{OD}$ where $A,B,C,D$ are the rectangle's summits and $O$ is the viewpoint.
When $O$ is exterior to the rectangle then there is a minimum and a maximum angle which determine a visibility sector.
The points associated with $\theta_{min}$ and $\theta_{max}$ are these you are searching for.
As Fabio Somenzi said, there are degenerated cases where two points can match, it's up to you to set a rule in this case.

Edit: 
There is in fact an issue in determining the angles, a condition for this to work is that $|\theta_i-\theta_j|\le\pi$ for all $i,j$.
One way to achieve this is to evaluate the shortest angle between two points, it is either the one measured clockwise or the one measured anticlockwise.
It happens we can achieve this with the formula defined in : 
Interpolating Between 2 Angles
And linked to a post on stack overflow

$\displaystyle f(\theta,\theta_0)=(((\theta-\theta_0)\mod {2\pi})+3\pi)\mod {2\pi})-\pi$


For instance assume a configuration like the one Fabio proposed. (I will use approximation in degrees here for simplicity of understanding)

We have the angles $\begin{cases} a=135° \\ b=153° \\ c=-153°\\ d=-135°\end{cases}$  
So the extrema here are $b,c$ but the points $B,C$ are obviously not the ones we are searching for.
But now if we transform the angles we get : 
For $\theta_0=a\quad \begin{cases} a'=f(a,a)=0° \\ b'=f(b,a)=18° \\ c'=f(c,a)=72°\\ d'=f(d,a)=90°\end{cases}\quad$ or for $\theta_0=c\quad \begin{cases} a'=f(a,c)=-72° \\ b'=f(b,c)=-54° \\ c'=f(c,c)=0°\\ d'=f(d,c)=18°\end{cases}$ 
Whichever reference $\theta_0$ we choose, the extrema are now always $a',d'$ and the points $A,D$ are effectively the visibility points.
A: A straight line/ray through A or B cuts the rectangle at two points. When the angle is increased away from rectangle's center so that it  cuts at one point only at corner, then then that point is the one you are searching for.
Repeat the process for another point.
In other words the three lines are *concurrent" at the corner vertex of rectangle.
But for visible cormers of rectangle sides are on one side of ray only but for the farther invisible point they are on either side of ray which cuts  through them.
A: As zwim writes in his answer, the vertices that correspond to the two extremes of the visual angles are the ones you want. However, you don’t really need to compute these angles explicitly. If you project the vertices onto a line that doesn’t pass through the view point, you can use the positions along the line of these projections as proxies for these angles: the vertices that have the outermost projections are the ones you want.  
A convenient choice for the line onto which to project is one that lies between the view point and the rectangle. This choice guarantees that none of the rays from the view point to a vertex is parallel to the line onto which you’re projecting, so you don’t have to deal with this additional (minor) complication. Since the rectangle is axis-aligned, you can always find either a horizontal or vertical line that meets this criterion, which simplifies the computations even further. I’ll illustrate with a vertical line, but the calculations for a horizontal line are similar.  

Given a view point $\mathbf p$ and line $\mathbf l$, the projection of a vertex $\mathbf v$ onto the line is simply the intersection of $\mathbf l$ with the line through $\mathbf p$ and $\mathbf v$. Working in homogeneous coordinates, this can be expressed via cross products as $\mathbf v'=\mathbf l\times(\mathbf p\times\mathbf v) = (\mathbf l^T \mathbf v)\mathbf p-(\mathbf l^T \mathbf p)\mathbf v = (\mathbf p\mathbf l^T-\mathbf p^T\mathbf l)\mathbf v$. Using the cross-product formula on the left is straightforward, but it might be less expensive to construct a matrix that captures the parenthesized term in the last expression and use that to compute the projections. Then, convert the resulting projection points $\mathbf v'$ to Cartesian coordinates and find the ones with the maximal and minimal $y$-coordinates.  
For the illustrated example, the view point is $(1,1)$, the four vertices of the rectangle are $(4,2)$, $(8,2)$, $(8,5)$ and $(4,5)$, and we will be projecting onto the line $x=2$. In homogeneous coordinates, $\mathbf l=[1:0:-2]$ and the homogeneous coordinates of all of the points are obtained by appending a $1$. We compute: $$\begin{align} [1:0:-2]\times([1:1:1]\times[4:2:1]) &= [6:4:3] \to \left(2,\frac43\right) \\
[1:0:-2]\times([1:1:1]\times[8:2:1]) &= [14:8:7] \to \left(2,\frac87\right) \\
[1:0:-2]\times([1:1:1]\times[8:5:1]) &= [14:11:7] \to \left(2,\frac{11}7\right) \\
[1:0:-2]\times([1:1:1]\times[4:5:1]) &= [6:7:3] \to \left(2,\frac73\right). \end{align}$$ The second and fourth vertices have the minimum and maximum $y$-coordinates, so those are the ones you want.  
In matrix form we have $$\mathbf p\mathbf l^T-\mathbf p^T\mathbf l\,I_3 = \begin{bmatrix}2&0&-2\\1&1&-2\\1&0&1\end{bmatrix}.$$ Since we’re only interested in the Cartesian $y$-coordinate of the projected point, we can discard the first row of this matrix. We can also fold in the conversion to Cartesian coordinates, yielding the formula $$y'={[1:1:-2]\cdot\mathbf v\over[1:0:1]\cdot\mathbf v}.$$ I’ll leave it to you to verify that this produces the same $y$-coordinates as the cross-product computation.  
If you end up with two vertices with the same max/min coordinate, you’ll need to compare their distances from the view point, which you can do with a simple coordinate comparison, and select the nearer one. If projecting onto a horizontal line instead, you would of course compare the $x$-coordinates instead of the $y$-coordinates of the projections.  
