In which direction does the normal force point if a rod can swivel? A rod is attached to a wall in such a way it can swivel. 
In this case: In which direction does the force (of the wall on the rod) point to? I drew the blue force as I would make a force diagram. Am I wrong?

Here is an example in which the rod can swivel, but now the normal force is perpendicular to the wall. The direction of the force here is different. Why? Is maybe one of the pictures wrong?

Also: What is the recipe here? How do we determine the direction?
 A: The pin forces can point in any direction, since all directions are constrained for motion.
You just can't have a normal force in the same direction as sliding is allowed because that would mean the joint can do/consume work.
Now for any example the actual direction is such that all forces converge to a single point. Slide the force vectors along their line of action such that they meet at a single point.

At this location the forces must balance, and that is how the magnitude and direction of the pin force (pink) is found, as well as the magnitude of the tension (black).
A: By definition, the normal force is always perpendicular to the common surface at which the objects make contact. Without a hinge, this surface would be the wall. There could also be friction - in which case the reaction force at the wall (ie the resultant of normal and friction forces) could point in any direction away from the wall. With a hinge, the normal force between it and the rod can point in any direction, even towards the wall.
There are two conditions for any object to be in static equilibrium :
1. The resultant force on it is zero. (Consequently, the components of force in each dimension add to zero.)
2. The resultant moment of forces about any point is also zero. (Consequently, the moments about each orthogonal rotation axis add to zero.)
These conditions can be used to find 2+1=3 unknowns in a 2D problem (in which all rotation axes are parallel), or 3+3=6 unknowns in a 3D problem. In n dimensions you need n components to define each force.
Condition 1 can be determined geometrically by drawing a polygon of forces, placing vectors representing the forces tail to head in any order. If the polygon is closed the resultant force is zero. Algebraically, the $x, y, z$ components of all forces separately sum to zero. 
Condition 2 is usually determined algebraically, by taking moments about any point in 2D or any axis in 3D. Often the hinge is most convenient, unless the hinge force is the one you need to find. 
Geometrically, if the lines of action of all forces pass through the same point then condition 2 is met. But the corollary (that if the forces don't intersect then condition 2 is not met) is only true for 3 or fewer forces, it is not necessarily true for 4 or more forces. I am not aware of any simple geometrical construction to cover the latter case.

In your 1st diagram, with the hinge reaction $F_1$ as shown, extending the line of action of each force it seems that $W, T, F_1$ do intersect at one point on the cable, so condition 2 is met. This is probably how the direction of $F_1$ was found - by joining the hinge to the intersection of $W$ and $T$. 
To find the magnitude of $F_1$ you still need to either :
A. Geometrically "close the polygon of forces" - ie draw vector $W$ and lines in the directions of $F_1$ and $T$ at either end of $W$, then their intersection gives the lengths - ie unknown magnitudes - of $F_1$ and $T$; or
B. Algebraically use the fact that the sums of vertical and horizontal components of all forces must both be zero.    
In your 2nd diagram there are 4 forces. The polygon of forces could be used to find either an unknown force (magnitude and direction) or two unknown magnitudes if directions are known (eg $F$ and $T$) using the construction in A above. 
However, closing the polygon of forces only satisfies condition 1. If the 4 forces intersect at a single point condition 2 would also be met; but it is obvious that they do not intersect in this example. In fact you can see from taking moments about the right-hand end of the rod that there is a resultant anti-clockwise moment of forces acting on it, assuming that the direction of $F$ is as given. The rod cannot be in static equilibrium under the given conditions. 
