I have studied on the internet that gases exert equal pressure in all directions in a container but liquids do not. [...] Why is that so?
Actually they both work in the same manner. The cause is the presence of gravity.
Pressure increases with depth in a liquid, because the heavy (dense) liquid has to carry the whole column of liquid above it. A water particle at the bottom of the sea must hold up all the water above it and all the air above that. The water particle at the surface only has to hold up the air above it (corresponds to standard atmospheric pressure).
It is the same thing for air and other gases. And as you might already know, the atmospheric pressure at ground level is much bigger than the atmospheric pressure at an air plane in a height of 10 km. Just watch any aircraft crash movie and see how everything is suched out when there is a breach because of the lower outside pressure...
For gas within an earth sized container, the pressure difference because of depth is so small because of the very low density that it simply doesn't have to be considered.
Also can there be a situation in which liquid can exert equal pressure on the walls of their container, independent of depth?
Yes, in outer space where no force like gravity pulls all particles in one single direction so they have to "carry" reach other. But in that case it would also be difficult to define depth...
I should mention though that such liquid in outer space of course exerts it's own gravitational pull. If you have large quantities of liquid (or gas for that matter - just look at a gas planet), and I mean very large quantities, then the liquid will form a sphere and the pressure will increase as you dive deeper. But this depth is then measured towards the center of this sphere.
Background
Let us assume we have a function, $f_{s}(\mathbf{x},\mathbf{v},t)$, which defines the number of particles of species $s$ in the following way:
$$
dN = f_{s}\left( \mathbf{x}, \mathbf{v}, t \right) \ d^{3}x \ d^{3}v
$$
which tells us that $f_{s}(\mathbf{x},\mathbf{v},t)$ is the particle distribution function of species $s$ that defines a probability density in phase space. We can define moments of the distribution function as expectation values of any dynamical function, $g(\mathbf{x},\mathbf{v})$, as:
$$
\langle g\left( \mathbf{x}, \mathbf{v} \right) \rangle = \frac{ 1 }{ N } \int d^{3}x \ d^{3}v \ g\left( \mathbf{x}, \mathbf{v} \right) \ f\left( \mathbf{x}, \mathbf{v}, t \right)
$$
where $\langle Q \rangle$ is the ensemble average of quantity $Q$.
Application
If we define a set of fluid moments with similar format to that of central moments, then we have:
$$
\text{number density [$\# \ (unit \ volume)^{-1}$]: } n_{s} = \int d^{3}v \ f_{s}\left( \mathbf{x}, \mathbf{v}, t \right) \\
\text{average or bulk velocity [$length \ (unit \ time)^{-1}$]: } \mathbf{U}_{s} = \frac{ 1 }{ n_{s} } \int d^{3}v \ \mathbf{v}\ f_{s}\left( \mathbf{x}, \mathbf{v}, t \right) \\
\text{kinetic energy density [$energy \ (unit \ volume)^{-1}$]: } W_{s} = \frac{ m_{s} }{ 2 } \int d^{3}v \ v^{2} \ f_{s}\left( \mathbf{x}, \mathbf{v}, t \right) \\
\text{pressure tensor [$energy \ (unit \ volume)^{-1}$]: } \mathbb{P}_{s} = m_{s} \int d^{3}v \ \left( \mathbf{v} - \mathbf{U}_{s} \right) \left( \mathbf{v} - \mathbf{U}_{s} \right) \ f_{s}\left( \mathbf{x}, \mathbf{v}, t \right) \\
\text{heat flux tensor [$energy \ flux \ (unit \ volume)^{-1}$]: } \left(\mathbb{Q}_{s}\right)_{i,j,k} = m_{s} \int d^{3}v \ \left( \mathbf{v} - \mathbf{U}_{s} \right)_{i} \left( \mathbf{v} - \mathbf{U}_{s} \right)_{j} \left( \mathbf{v} - \mathbf{U}_{s} \right)_{k} \ f_{s}\left( \mathbf{x}, \mathbf{v}, t \right) \\
\text{etc.}
$$
where $m_{s}$ is the particle mass of species $s$, the product of $\mathbf{A} \mathbf{B}$ is a dyadic product, not to be confused with the dot product, and a flux is simply a quantity multiplied by a velocity (from just dimensional analysis and practical use in continuity equations).
In an ideal gas we can relate the pressure to the temperature through:
$$
\langle T_{s} \rangle = \frac{ 1 }{ 3 } Tr\left[ \frac{ \mathbb{P}_{s} }{ n_{s} k_{B} } \right]
$$
where $Tr\left[ \right]$ is the trace operator and $k_{B}$ is the Boltzmann constant. In a more general sense, the temperature can be (loosely) thought of as a sort of pseudotensor related to the pressure when normalized properly (i.e., by the density).
Answers
How can a Hot gas be Low Pressured?
If you look at the relationship between pressure and temperature I described above, then you can see that for low scalar values of $P_{s}$, even smaller values of $n_{s}$ can lead to large $T_{s}$. Thus, you can have a very hot, very tenuous gas that exerts effectively no pressure on a container. Remember, it's not just the speed of one collision, but the collective collisions of the particles that matters. If you gave a single particle the enough energy to impose the same effective momentum transfer on a wall as $10^{23}$ particles at much lower energies, it would not bounce off the wall but rather tear through it!
How can a High Pressured gas be Cold?
Similar to the previous answer, if we have large scalar values of $P_{s}$ and even larger values of $n_{s}$, then one can have small $T_{s}$. Again, from the previous answer I stated it is the collective effect of all the particles on the wall, not just the individual particles. So even though each particle may have a small kinetic energy, if you have $10^{23}$ hitting a wall all at once, the net effect can be large.
Best Answer
The pressure of a gas is defined as the force the gas would exert upon a surface or container. However, there is no need for a container for pressure to exist. For instance, the air you're breathing right now (unless you're in an airplane or submarine) has pressure due to the column of atmosphere above you. Stars are balls of gas (plasma, actually) that are pressurized by gravity; no containers to be seen.