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\begin{questionparts}
\item Notice that if $x > 0$ we must have
\begin{align*}
&& \left ( \sqrt{x} - \frac{1}{\sqrt{x}} \right)^2 &\geq 0 \\
\Leftrightarrow && x - 2 + x^{-1} & \geq 0 \\
\Leftrightarrow && x + x^{-1} & \geq 2
\end{align*}

Let $S$ be the sum, and assume all probabilities are equal
\begin{align*}
&& \mathbb{P}(S = 2) &= q_1 r_1 \\
&& \mathbb{P}(S = 12) &= q_6 r_6 \\
&& \mathbb{P}(S = 7) &= \sum_{i=1}^6 q_i r_{7-i} \\
\Rightarrow && q_1r_1 &= q_6r_6 \\
\Rightarrow && q_1r_6+q_6r_1 &\leq  q_1r_1 \\
\Rightarrow && \frac{r_6}{r_1} + \frac{q_6}{q_1} &\leq 1 \\
\Rightarrow && q_1r_6+q_6r_1 &\leq  q_6r_6 \\
\Rightarrow && \frac{q_1}{q_6} + \frac{r_1}{r_6} &\leq 1 \\
\Rightarrow && \frac{r_6}{r_1} + \frac{q_6}{q_1}+\frac{q_1}{q_6} + \frac{r_1}{r_6} &\leq 2\\
\text{but} &&  \frac{r_6}{r_1} + \frac{q_6}{q_1}+\frac{q_1}{q_6} + \frac{r_1}{r_6} &\geq 4
\end{align*}

Since we have a contradiction they cannot all be equal.

\item We would like $\displaystyle \sum q_i^2 \leq 1/36$ (subject to $\displaystyle \sum q_i = 1$, clearly this cannot be true since:

\begin{align*}
&& 1 &= \left ( \sum_{i=1}^6 q_i \right)^2 \\
&&&=  \sum_{i=1}^6 q_i^2 + \sum_{i \neq j} 2q_i q_j \\
&&&\leq  \sum_{i=1}^6 q_i^2 +  5\sum_{i=1}^6 q_i^2 \\
&&&=6 \sum_{i=1}^6 q_i^2 \\
\Rightarrow &&  \sum_{i=1}^6 q_i^2 &\geq 1/6 > 1/36
\end{align*}

[For a weaker solution to the last part, notice that the largest value of $q_i$ is  $\geq 1/6$ and therefore $q_{max}^2 \geq 1/36$, but if equality holds then the other values must also be non-zero, and therefore the inequality cannot hold]
\end{questionparts}

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