This paradox arises due to wrong application of Newton's second and third laws. Before applying Newton's laws, we should decide 'what is the system?'. Once we identify the 'system', then it is possible to identify all the forces acting on the system. We should not consider the force exerted by the system. If there is an unbalanced force acting on the system, then it should have acceleration in the direction of the resultant force. By following these steps we will analyse the horse and cart motion.
If we decide on the cart+horse as a 'system', then we should not consider the force exerted by the horse on the cart or the force exerted by cart on the horse. Both are internal forces acting on each other. According to Newton's third law, total internal force acting on the system is zero and it cannot accelerate the system. The acceleration of the system is caused by some external force. In this case, the force exerted by the road on the system is the external force acting on the system. It is wrong to conclude that the total force acting on the system (cart+horse) is zero without including all the forces acting on the system. The road is pushing the horse and cart forward with acceleration. As there is an external- force acting on the system, Newton's second law has to be applied and not Newton's third law. The following figures illustrates this.

If we consider the horse as the 'system', then there are three forces acting on the horse.
(i) Downward gravitational force (m_hg)
(ii) Force exerted by the road (F_r)
(iii) Backward force exerted by the cart (F_c)
It is shown in the following figure.

F_r - Force exerted by the road on the horse
F_c - Force exerted by the cart on the horse
F^丄_r-Perpendicular component of F_r=N
F^||_r-Parallel component of Fr which is reason for forward movement.
The force exerted by the road can be resolved into parallel and perpendicular components. The perpendicular component balances the downward gravitational force. There is parallel component along the forward direction. It is greater than the backward force (F_c). So there is net force along the forward direction which causes the forward movement of the horse.
If we take the cart as the system, then there are three forces acting on the cart.
(i) Downward gravitational force (m_cg)
(ii) Force exerted by the road (F_r)
(iii) Force exerted by the horse (F_h)
It is shown in the figure.

The force exerted by the road (\(\vec { { F }_{ r } } \) ) can be resolved into parallel and perpendicular components. The perpendicular component cancels the downward gravity (m_cg). Parallel component acts backwards and the force exerted by the horse (\(\vec { { F }_{ h } } \) ) acts forward. Force (\(\vec { { F }_{ h } } \) ) is greater than the parallel component acting in the opposite direction. So there is an overall unbalanced force in the forward direction which causes the cart to accelerate forward.
If we take the cart+horse as a system, then there are two forces acting on the system.
(i) Downward gravitational force (m_h + m_c)g
(ii) The force exerted by the road (F_r) on the system.
It is shown in the following figure.

(iii) In this case the force exerted by the road (F_r) on the system (cart+horse) is resolved in to parallel and perpendicular components. The perpendicular component is the normal force which cancels the downward gravitational force (m_h + m_c)g. The parallel component of the force is not balanced, hence the system (cart+horse) accelerates and moves forward due to this force.

In a leveled circular road, skidding mainly depends on the coefficient of static friction \({ \mu }_{ s }\). The coefficient of static friction depends on the nature of the surface which has a maximum limiting value. If the speed of car exceeds this safe speed, then it starts to skid outward but frictional force comes into effect and provides an additional centripetal force to prevent the outward skidding. 
\(tan \ \theta> { \mu }_{ s }\)
When the tangent of the angle of banking is greater than the coefficient of friction, skidding occurs.

When the block slides, the force acting on the block is kinetic friction which is equal to f_k= μ_smg.
From Newton's second law ma = -μ_smg
The negative sign implies that force acts on the opposite direction of motion.
The acceleration of the block while sliding a =-μ_kg
The negative sign implies that the acceleration is in opposite direction of the velocity. Note that the acceleration depends only on g and the coefficient of kinetic friction μ_k. We can apply the following kinematic equation
v=u+at
The final velocity is zero
0=u-u_kgt
t=\(\frac { u }{ { \mu }_{ k }g } \).

Note: Before applying Newton's laws, the following steps have to be followed:
1. Choose a suitable inertial coordinate system to analyse the problem. For most of the cases we can take Earth as an inertial coordinate system.
2. Identify the system to which Newton's laws need to be applied. The system can be a single object or more than one object.
3. Draw the free body diagram.
4. Once the forces acting on the system are identified, and the free body diagram is drawn, apply Newton's second law. In the left hand side of the equation, write the forces acting on the system in vector notation and equate it to the right hand side of equation which is the product of mass and acceleration. Here, acceleration should also be in vector notation.
5. If acceleration is given, the force can be calculated. If the force is given, acceleration can be calculated.

By following the above steps:
We fix the inertial coordinate system. on the ground as shown in the figure.
The forces acting on the mango are
(i) Gravitational force exerted by the Earth on the mango acting downward along negative y-axis.
(ii) Tension (in the cord attached to the mango) acts upward along positive y-axis.
The free body diagram for the mango is shown in the figure


\(\bar { { F }_{ g } } =mg(-\hat { j } )=-mg\hat { j } \)
Here, mg is the magnitude of the gravitational force and \((\hat { -j } )\) represents the unit vector in negative y-direction.
\(\vec { T } =T\hat { j } \)
Here T is the magnitude of the tension force and \((\hat { j } )\) represents the unit vector in positive y direction.
\(\vec { F_{ net } } ={ F }_{ g }+\vec { T } =-mg\hat { j } +T\hat { j } =(T-mg)\hat { j } \)
From Newton's second law \(\vec { F_{ net } } =m\vec { a } \)

Since the mango is at rest with respect to us (inertial coordinate system) the acceleration is zero \((\vec { a } =0)\)
So, \(\vec { F_{ net } } =m\vec { a } \) = 0
\((T-mg)\hat { j } \) = 0
By comparing the components on both sides of the above equation, we get T - mg = 0. So the tension force acting on the mango is given by T = mg.
Mass of the mango m = 400 g and g = 9.8 ms^-2
Tension acting on the mango is T = 0.4\(\times\)9.8 = 3.92 N.

Forces on Block A:
(i) Downward gravitational force exerted by the Earth (m_Ag)
(ii) Upward normal force (N_B) exerted by block B (N_B)
The free body diagram for block A is as shown in the following picture.

Forces on Block B:
(i) The downward gravitational force exerted by Earth (m_Bg)
(ii) The downward force exerted by block A (N_A)
(iii) An upward normal force exerted by block C(N_c)

Forces on Block C:
(i) Downward gravitational force exerted by Earth (m_Cg)
(ii) Downward force exerted by block B (N_B)
(iii) Upward force exerted by the table (N_table)