Is tension a conservative force? Are forces such as tension (from an in extensible string), normal reaction, and applied force from us, non conservative forces? If so why?
I have read in few books that these forces are labeled as nonconservative, but most of the time they are internal and do zero work and thus mechanical energy of the system remains constant, shouldn't they be conservative? 
 A: First, the definition of conservative force is not "a force that does no work".
Second, you can always make your system large enough so that total mechanical energy is conserved for the system, so this cannot determine whether forces are conservative or not either. Internal vs. external is a subjective distinction, whereas conservative vs. nonconservative is an objective distinction.
Finally, tension, normal forces, etc. are tricky, since they do different things in different scenarios. You have to be more specific about the specific scenario to discuss those in more detail. But in general they are not conservative, as the work done by forces like these can depend on the the path taken, or how the same path is "traversed".

consider a block man system, let the man pull the block towards left by a constant force say F, if another external source makes a round trip for the block, dosent it mean that the work done by the man is zero which implies it is conservative?

That's what I meant by it being tricky. Constant forces are conservative, and mathematically there isn't anything distinguishing the pulling force and an actual force that is always constant. However, in reality pulling is more complicated. You can pull more or less. You could pull the same object the same distance but faster, thus the work done by your pulling doesn't just depend on the end points of the path, i.e. the force isn't actually conservative. A "pulling force" isn't really a fundamental force (at least macroscopically), and it would be even weirder to say you are storing potential energy by pulling on the block. 
I don't fully follow your logic about an additional external force. I think you need to clarify that point.
A: First, you need to know how a conservative force is defined. A force is conservative if the work done by it is path independent.
Now coming to your doubt, according to the work-energy theorem,
$$
\text{work (done by all the forces} = \Delta\left(\text{kinetic energy}\right).
$$
If you know, potential energy is defined only for conservative forces so,
$$
\text{work (done by a conservative force)} = -\Delta U,
$$
the negative of the change in potential energy.

So, take an example where tension and gravity (conservative) is acting on a pendulum. Applying the work-energy theorem,
$$
\text{work (done by tension)} + \text{work (done by gravity)} = \mathrm{K.E.} \text{ (final)} - \mathrm{K.E.} \text{ (initial)}
\tag{1}
$$
which implies
\begin{align}
\text{work (gravity)} & = - \left( U\text{ (final)} - U \text{ (initial)}\right)
\\
\text{work (tension)} & = 0.
\end{align}
Putting these in $(1)$ we get
\begin{align}
\mathrm{K.E.} \text{ (final)} + U \text{ (final)} = \mathrm{K.E.} \text{ (initial)} + U \text{ (initial)},
\end{align}
i.e., mechanical energy is conserved. Now, if the tension had done work this equation would be invalid and the mechanical energy would not have been conserved.
Observe that even though gravity (conservative force) is doing work the total mechanical energy is conserved but if the tension had done any work the mechanical energy of the system wouldn't have been conserved. This equation of conservation of mechanical energy is only valid if the work done by tension is zero, and is invalid if tension starts doing any work, hence a non conservative force.
