And: If you have a book on a table the book is exerted a force on the table (weight due to gravity),
That's where you went wrong. The force that the book exerts on the table is not a gravitational force, it's a normal force.
and the table reacts with an equal and opposite force.
That's also a normal force. So the book exerts a (normal) force on the table, and the table exerts a (normal) force on the book.
But the force acting on the table is due to gravity (is this the same as a gravitational force?),
No, it's not, and in fact this force (the normal force) is only indirectly due to gravity. The only relevant gravitational force is the force exerted by the Earth on the book. And the book also exerts a gravitational force back on the Earth, but because the Earth is so heavy, that force has no noticeable effect. (The Earth also exerts a gravitational force on the table, and the table on the Earth, but those don't matter so much in this particular scenario.)
As usual for mechanical statics problems, the entire thing becomes clear if you draw a free body diagram.
Here we have the object (red) sitting on top of a table (blue) which is sitting on the Earth (black curved line).
The object experiences two forces
It's weight $W_o$ which is caused by gravitational interaction with the Earth.
A normal force $N_{o,T}$ from the table top. Read the subscript $o,T$ as the force "on the object from the Table".
This probably already answers your question, but let's go on a bit.
The "reaction" force to $W_o$ is an equally strong gravitational pull acting on the Earth itself.
The reaction to the normal force is an equally strong normal force pushing down on the table.
The table also experiences a normal force $N_{T,o}$ from the object.
This is the reaction to $N_{o,T}$ and has equal magnitude to $N_{o,T}$.
the table of course also has a weight force $W_T$ and the Earth feels a reaction as shown.
Because the system is static we know that the forces on the object are balanced
$$N_{o,T}=W_o \, .$$
We also know that the forces on the table must be balanced
$$N_{T,E} = W_T + N_{T,o} \, .$$
As we said, the normal force on the table from the object is a reaction to the normal force on the object from the table, so their mangitudes are equal,
$$N_{T,o} = N_{o,T} \, .$$
Therefore
$$N_{T,E} = W_T + N_{o,T} = W_T + W_o \, .$$
This is just a careful derivation of the fact that the upward normal force on the table from the Earth must be the sum of the weight of the table and the object!
It's just saying that if you pick up the table you feel the weight of both the table and the object on the table.
Best Answer
This is correct.
This is also correct. The force of gravity and the normal force from the table on the ball are not action-reaction pairs as described by Newton's third law.
As you have recognized, both forces in action-reaction force pairs do not act on the same object, so why would you think that in order to have forces acting on an object they have to be part of an action-reaction pair?
There are two action-reaction force pairs here.
As you can see, we have our two forces acting on our ball: gravity and normal force. Each of these forces has a corresponding force that it forms an action-reaction pair with, as stated by Newton's third law.
As a small aside, the more I see questions on this site about Newton's third law and actions/reactions, the more I realize how confusing this terminology is to new students to physics. If this is all still confusing to you, may I suggest a different view of Newton's third law?
If you want to pick out your action-reaction pairs, just pick out your interactions. Are the ball and Earth interacting? Yes, through gravity. Are the ball and the table interacting? Yes, through the normal force (electrostatic interactions).