Is hydrostatic pressure independent of temperature? Is hydrostatic pressure independent of temperature?
Why only the weight of overlaying liquid is taken into account for calculating hydrostatic pressure and not the collision forces due to temperature?
In space (if temperature is not zero),will the liquid pressure on walls be zero?
 A: The effect of temperature is included in determining the density as a function of temperature and pressure via the equation of state.  For example, in the case of an ideal gas in hydrostatic equilibrium, the density is related at each point in the gas locally to the temperature and pressure by $$\rho=\frac{pM}{RT}$$where M is the molar mass. So the ideal gas law is satisfied at each point locally within the gas.  The static equilibrium equation is then $$\frac{dp}{dz}=\rho g = \frac{pM}{RT}g$$where z is the depth.  So the temperature and the equation of state are taken into account.  For a liquid, we get$$\frac{dp}{dz}=\rho (T,p)g$$which also includes the equation of state $\rho=\rho(T,p).$
A: Yes, it is independent in the case to which I believe you are referring, in which one has a container of liquid which is open to the atmosphere here on Earth, and the system is in static equilibrium.  However these conditions are merely a special case in which the various effects of temperature cancel each other out perfectly.
There are many possible scenarios in which one might discuss the pressure exerted by a fluid.  The temperature of fluid is relevant in many of them:  In the case of a gas, one can often use a mental model in which the molecules are alternately undergoing free ballistic motion and brief collision events.  In a closed container, temperature is the main factor in determining the pressure.  If given an arbitrary density distribution of a gas at some moment and asked to find the pressure distribution, one could use the temperature and the molar mass of the gas to directly calculate the pressure from the frequency of collisions and the average collision momentum.
However, the case of a static system, which is open to the atmosphere, is different.  Since the system is static, we know that the fluid is not flowing or accelerating anywhere, so the net force on any chunk of fluid is zero.  The net force on any such chunk consists of an integral of pressure forces around the boundary of the chunk plus the gravitational force on the chunk.  There are no other forces except for the constant atmospheric pressure boundary condition at the container opening, so the pressure distribution necessary for static equilibrium is completely determined by the weight distribution of the fluid.  (If the container were closed, then the boundary conditions on the pressure could be temperature dependent.)
This does not change the answer to your question, but I would also like to clarify that for a liquid, your mental model of ballistic molecule motion with distinct collision events is inappropriate.  The molecules of a liquid are always loosely bound to each other, and the pressure of the liquid is not simply computed from collision frequency and temperature.
