. . . system?¹

 There must be a limit to the amount of energy the system can hold, otherwise more energy is bound to offer more possibilities for redistribution, and thus more entropy. If the entropy is decreasing with increasing energy (i.e. the derivative is negative) then by definition the temperature is negative (i.e. hotter than infinite temperature.)

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. . . zero. ²

 There is only one point where the electric field is zero: exactly as far to the right of the negative charge as the separation between the charges.

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. . . temperature.³

 The probability of a given single particle state of a given energy being occupied is the same for both, but the density of states (i.e. the number of possible states with speeds within a given $dv$ of $v$) is different because there are more directions for the vector velocity to point in 3 dimensions than in 1 dimension. You can also describe this in terms of modes of standing waves, but the classical explanation is adequate for full credit.

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. . . sentences:⁴

 The electric field inside the tube ($r<R$) is a complicated function of the charge's position. The electric field outside the tube ($r>R$) has a magnitude $E \approx Q/2\pi\epsilon_\circ L r$ (except near the ends) and is in the $\hat{\mbox{\boldmath$r$\unboldmath}}$ direction. This is because a net charge $-Q$ is distributed over the inside of the conductor in such a (complicated) way as to terminate all the electric field lines emitted by the point charge so that the field can be zero inside the conductor, leaving an equal positive charge $Q$ to be distributed uniformly over the outer surface of the conductor. So an observer at $r>R$ "sees" a uniformly charged cylinder with a net charge per unit length of $\lambda = Q/L$, producing an electric field $\vec{\mbox{\boldmath$E$\unboldmath}} = \hat{\mbox{\boldmath$r$\unboldmath}} \lambda/2\pi r$.

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. . . charge?⁵

 To the left of the positive sheet of charge and to the right of the negative sheet of charge, the electric field is zero, because a Gaussian "pillbox" surface cutting through the entire array at right angles encloses no net charge.

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. . . slab?⁶

 A negative surface charge is attracted to the left side of the conductor by the positive sheet of charge, leaving behind an equal and opposite positive surface charge on the right side. These induced surface charges must cancel the electric fields of the sheets inside the conductor, so they produce electric fields of their own with the same magnitude as those of the sheets, and (between the sheets and the conductor) in the same direction. So in both gaps we get an electric field to the right whose (uniform) magnitude is $E_{\rm gap} = \sigma_\circ/\epsilon_\circ$ , twice that from a single sheet, $E_\circ = \sigma_\circ/2\epsilon_\circ$.

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. . . slab?⁷

 The electric field inside any conductor is zero.

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. . . . ⁸

 There are several ways to think about this question; the trick is not to get them mixed up. The reasoning given in the answer to the previous question treats all sheets of charge (original or induced) on an equal basis, explaining the zero electric field within the conductor explicitly in terms of the sum of several contributions. This immediately gives $\vert\sigma\vert = \sigma_\circ$ on both sides of the slab. You can also use the general rule (obtained by taking a very small Gaussian "pillbox" that only encloses one surface of the conductor) that $E = \sigma/\epsilon_\circ$ near the surface of any conductor; in this case $E_{\rm gap} = \sigma/\epsilon_\circ$ but since $E_{\rm gap} = 2 E_\circ$ and $E_\circ = \sigma_\circ/2\epsilon_\circ$ we get $\sigma/\epsilon_\circ = 2 \sigma_\circ/2\epsilon_\circ$ or $\vert\sigma\vert = \sigma_\circ$ again.

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. . . Explain. ⁹

  Zero . There is no net charge enclosed within a coaxial Gaussian cylinder of radius $r < a$.

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. . . Explain. ¹⁰

  Zero . There is a positive charge $\lambda \ell$ from the inner conductor and a negative charge $-\lambda \ell$ from the outer conductor both enclosed within a coaxial Gaussian cylinder of length $\ell$ and radius $r > b$, but there is still no net charge enclosed.

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. . . ? ¹¹

 Now it gets interesting. A coaxial Gaussian cylinder of length $\ell$ and radius $r$ between $a$ and $b$ encloses a net positive charge $Q = \ell \lambda$ which must equal $\epsilon_\circ E$ times the surface area $2\pi r \ell$ through which $\vec{\mbox{\boldmath$E$\unboldmath}}$ emerges, giving the familiar $E(r) = \lambda / 2\pi \epsilon_\circ r$ . The negative shell of charge at $r=b$ is not enclosed, so it contributes nothing to $E$.

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. . . ? ¹²

 You are welcome to do an integral if you wish, but I excused you from this chore by specifying that you can treat $E$ as constant in the narrow gap, giving simply $\varepsilon = e E (b - a)$. Check the sign: the negatively charged electron experiences an inward force so "out" is "uphill" and $U(b)$ is greater than $U(a)$. With $E \approx E(a) = \lambda / 2\pi \epsilon_\circ a$ this gives $\varepsilon = e \lambda (b - a) / 2\pi \epsilon_\circ a$. Putting in the numbers gives $\varepsilon = {(1.602 \times 10^{-19})(10^{-10})(0.01) \over (2\pi)(0.8854 \times 10^{-11})(1)}$ or $\varepsilon = 0.288 \times 10^{-20}$ J .

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. . . surface?¹³

 Here we have two states of different energies and therefore different Boltzmann factors. If we define the energy of a bead stuck to the surface at $r=a$ to be zero, then the ratio of the probability $P_b$ of being stuck to the surface at $r=b$ to the probability $P_a$ of being stuck to the surface at $r=a$ is $\exp(-\varepsilon/k_{_{\rm B}}T)$, where $T = 300$ K and $k_{_{\rm B}} = 1.3807 \times 10^{-23}$ J/K, giving $\varepsilon/k_{_{\rm B}}T = {0.288 \times 10^{-20} \over 0.4142 \times 10^{-20}} = 0.6953$ and $P_b/P_a = e^{-0.6953} = 0.499 \approx 1/2$. Since $P_a + P_b = 1 \approx (3/2)P_a$, $P_a \approx 2/3$ and $P_b \approx 1/3$, so on average 67 of the 100 beads will be stuck to the surface at $r=a$ and 33 will be stuck to the surface at $r=b$.

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