How do slight changes in these properties result in a large change in pressure, microscopically?
Slight change of volume is not so easy to accomplish for solids - it takes a great force to achieve it. Considerable external force applied by different body (wall) needs to be maintained. The pressure is a measure of this force per unit area and since the force is great, also the pressure is great.
One way to explain this is the following. In solids, the atoms are so close to each other that the inter-atomic repulsion forces become close to comparable to intra-atomic forces between charged particles that constitute them.
The electric force between particles gets stronger at small distances, since the Coulomb law says it changes with distance of particles $r$ as $1/r^2$. To change the volume of a solid appreciably, the force per area associated with one atom would have to be comparable to the Coulomb electric force of the nucleus on the electron. Its value for atoms is so immense that it is unattainable by common standards. It would be also hard to find a container that would keep its shape and resist expansion under such great forces. Hence changes in volume by forces of common magnitude are commonly negligible.
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
As the comments to the question have stated, in real gasses ( contrasted to ideal gasses which just bounce around elastically) there exist both elastic and inelastic scatterings controlled by quantum mechanical interactions.
Photons are generated leading to what we call Black Body radiation and an isolated gas volume will lose energy according to the Stephan Boltzmann law.
Thus the gas does lose energy if the temperature of matter surrounding it is lower.
In answer to
Elastic means an interaction of two particles where before and after , kinetic energy is conserved. If one assumes that only kinetic energies exist for this scatter ( as in the ideal gas) then energy is conserved because what one particle loses the other gains . If there are other forms of energy that can contribute to the two particle interaction then it is the total energy that is conserved. With billiard balls classically friction has to be taken into account with the energy balance, the same with the bouncing ball, and the kinetic energies stop being the total energy of the system. For particles in a gas it is the quantum mechanical framework, described above.