1."If there are other physical quantities available why work done is needed. what is special about work done that other quantities can not give us."
Work done is mathematically defined as the scalar product of the force $\vec F$ and the displacement $\vec s$. So
$W=\vec F.\vec s$ for constant forces. In case of variable forces we say something similar(I have just broken down the scalar product):$$W=\int F.dx +\int F.dy +\int F.dz$$
Here I am basically describing that total work done on an object is the sum of the work done by the individual forces acting on it resolved in coordinate axes of your choice.
So, what is special about work done that cannot be described by other physical quantities?
Work done as you might have guessed provides a relation between a force and the 'displacement of the object'(not necessarily caused by the force itself). That's why, we take the $\vec F$ and 'scale' it(stretching or compressing) to the $\vec s$ (or vice versa, but this makes more sense). You can imagine that we do this to show: just exactly 'how much' does this force contribute to the change in position of the object. We don't care about how fast the object changes its position(that's power), we just want to know what this force is doing in a system.
You can check, that no other physical quantities give us this relation, and the reason why it's needed in the first place is that: All forces contribute to net acceleration(that's what it means to be a force), but having information about its relation with displacement can tell us if individual forces are 'taking away' from the system, or 'putting something into' the system; this can be explained by the principle of energy, which tells us, not just about the current state of an object, but also about how this object will behave and interact with other objects in the future--bonus points because it is conserved all over the universe with NO EXCEPTIONS, it might as well be a base quantity!
$(A)$Hopefully I have described the purpose of $F$ and $d$ in the formula. In example (a), there are multiple forces acting on the system: Tension, gravitation, possibly friction, normal force by the box. You are correct, applied force at an angle does cause displacement; but does all of it goes into displacing the object? Of course not! Clearly a part of it goes into countering gravity and part of it is parallel. Is the part countering gravity doing work? No. This is not just because $cos\theta$ is $0$ at $\pi/2$ radians, but because it just makes sense! This part of the force neither takes away from the system, nor puts in anything. Think this one through.
$(B)$In part (b), we use the aforementioned reasoning again! Gravitational force field is indeed the only force acting on the system during the projectile motion, but its initial upward displacement comes from the external force applied by us. Therefore as it is going upwards(and gravity is slowing it down), gravitational force field is taking away from the system; the $\vec s$ and $\vec F_g$ are opposite in direction; which implies, work done by gravity is negative for the first part of the motion. But as it comes down, gravity contributes to motion, and work done is positive!
$(C)$In part (c), the exact same reasoning can be used. I will let you work this one out.
HINT: Once again, an applied force and a field force is acting on the system. Therefore, individual work done and total work done will be different.