It will actually weigh $F_b+\rho_lV_lg$ which is $\rho_lV_bg+\rho_lV_lg$, and not one of the two options you say.
Consider the liquid as the system and the block as an external body. Now we know that the liquid applies a buoyant force on the block. According to Newton's third law, the block will apply a reaction force on the liquid, equal in magnitude and opposite in direction. Thus total force on liquid is $F_b+F_g$ which gives $\rho_lV_bg+\rho_lV_lg$.
Note that this is the weight when the block is attached to the spring balance, and suspended in the liquid. If the block is kept on the floor not attached to the spring balance, the reading weight will be different.
Edit:
In the case when the block is resting on the floor, the weight will simply be $\rho_lV_bg+\rho_lV_lg$, because when you consider the block and liquid as a system, the buoyant force will become an internal force and cancel out on the whole system. So the only force responsible for the weight will be gravity.
In the elevator scenario, the elevator frame is getting accelerated; hence, the when you draw the free-body diagram, with respect to the elevator, the pseudo force acts downwards (opposite to the direction in which the frame is getting accelerated). Hence, the apparent weight increases as the pseudo force gets added up with the weight of the person.
Suppose the acceleration of the elevator is $a$ and the mass of the body is $m$, then the apparent weight of the body in the elevator frame is -
$$
N = m(a + g)
$$
In the second scenario, the buoyant force acts in the upward direction, because the buoyant force is always directed against the pressure gradient i.e, the direction in which pressure decreases. (Much like an electric field directed in the direction where the potential decreases)
Of course, the buoyant force exerted is equal to the weight of the fluid displaced by the body (Which is the Archimedes principle); but -
Drawing the FBD in the second case yields the weight of the body acting downwards, and the buoyant force acting upwards. This results in the weight decreasing (since the buoyant force is subtracted from the weight, not added up with it), and not increasing.
Say, the buoyant force acting on the body is $B$ and the actual weight is $W$, the net weight of the body (acting in the downward direction) then would be -
$$
W' = W - B
$$
Which is why the apparent weight of the body in the liquid decreases.
(This is considering that the density of the body is greater than the density of the liquid, in the case where it is opposite (the body doesn't sink; but floats partially), the signs of $W$ and $B$ are swapped and the net force is acting in the upward direction. In another scenario where the the weight of the the body is equal to the buoyant force, the net force on the body then is zero, hence it floats being completely submerged)
Keep in mind that a body loses weight in a liquid which is equal to the weight of the liquid displaced by it/equal to the buoyant force.
As for the bonus question, look into the answer to this question -
https://physics.stackexchange.com/a/296537/134658
Best Answer
A normal force is simply a force that arises from contact between one object and another object or substance. There can be several normal forces acting on an object. For example, an object on a slope held in place by a wedge will have two normal forces acting on it - one from the slope, the other from the wedge. Neither one is equal to the weight of the object (or its apparent weight if the slope is in an accelerating lift).
Apparent weight is a very specific instance of a normal force - it is the normal force registered by horizontal scales on which the object rests. If the scales and the object are in an accelerating lift then the apparent weight will be greater than or less than the “true” (unaccelerated) weight of the object, depending on the acceleration of the lift.
Buoyancy is another type of normal force which acts on objects partly or fully submerged in a liquid. But it is not the same as apparent weight. If you take a submerged object that is denser than the liquid and rest it on unaccelerated horizontal scales, the normal force registered by the scales will be equal to the object’s weight in air minus the buoyancy force. This is the apparent weight of the (unaccelerated) object when submerged in the liquid.
If the scales (still in the liquid) are accelerated up or down then the apparent weight of the object (as registered by the scales) will increase or decrease. But the buoyancy force stays the same. So this also confirms that buoyancy and apparent weight are not the same - they are different instances of normal forces.