Hint:
Write,
$$ \tag{1}\textstyle
P[\,X>31\,] =P\bigl[\,Z>{31-\mu\over\sigma}\,\bigr]=.2743\Rightarrow {31-\mu\over\sigma} = z_1
$$
$$\tag{2}\textstyle
P[\,X<39\,] =P\bigl[\,Z<{39-\mu\over\sigma}\,\bigr]=.9192\Rightarrow {39-\mu\over\sigma} =z_2 ,
$$
where $Z$ is the standard normal random variable.
You can find the two values $z_1$ and $z_2$ from a cdf table for the standard normal distribution. Then you'll have two equations in two unknowns. Solve those for $\mu$ and $\sigma$.
For example, to find $z_1$ and $z_2$, you can use the calculator here. It gives the value $z$ such that $P[Z<z]=a$, where you input $a$.
To use the calculator for the first equation first write
$$\textstyle P\bigl[\,Z<\underbrace{31-\mu\over\sigma}_{z_1}\,\bigr]=1-P\bigl[\,Z>{31-\mu\over\sigma}\,\bigr] =1-.2743=.7257.$$
You input $a=.7257$, and it returns $z_1\approx.59986$.
To use the calculator for the second equation,
$$\textstyle P\bigl[\,Z<\underbrace{39-\mu\over\sigma}_{z_2}\,\bigr]= .9192,$$
input $a=.9192$, the calculator returns $z_2\approx1.3997$.
So, you have to solve the system of equations:
$$
\eqalign{
{31-\mu\over\sigma}&=.59986\cr
{39-\mu\over\sigma}&=1.3997\cr
}
$$
(The solution is $\sigma\approx 10$, $\mu\approx 25$.)
(a) Yes. If $X \sim \operatorname{Normal}(\mu, \sigma^2)$, then the PDF of $X$ is given by $$f_X(x) = \frac{1}{\sqrt{2\pi}\sigma} e^{-(x-\mu)^2/(2\sigma^2)}, \quad -\infty < x < \infty.$$ We can also readily observe that $X$ is a location-scale transformation of a standard normal random variable $Z \sim \operatorname{Normal}(0,1)$, namely $$X = \sigma Z + \mu,$$ or equivalently, $$Z = \frac{X - \mu}{\sigma},$$ and the density of $Z$ is simply $$f_Z(z) = \frac{1}{\sqrt{2\pi}} e^{-z^2/2}, \quad -\infty < z < \infty.$$ Therefore, if $m$ is the median of $X$, then the median of $Z$ is $m^* = (m - \mu)/\sigma$. But we also know that $m^*$ satisfies $$F_Z(m^*) = \int_{z=-\infty}^{m^*} f_Z(z) \, dz = \Phi(m^*) = \frac{1}{2}.$$ But since $f_Z(z) = f_Z(-z)$ for all $z$, the substitution $$u = -z, \quad du = -dz$$ readily yields $$F_Z(m^*) = -\int_{u=\infty}^{-m^*} f_Z(-u) \, du = \int_{u=-m^*}^\infty f_Z(u) \, du = 1 - F_Z(-m^*),$$ and since both of these must equal $1/2$, we conclude $F_Z(m^*) = F_Z(-m^*)$, or $m^* = -m^*$, or $m^* = 0$. From this, we recover the median of $X$: $m = \sigma m^* + \mu = \mu$.
(b) The interquartile range is equal to $q_3 - q_1$, where $q_3$ satisfies $F_X(q_3) = \frac{3}{4}$ and $F_X(q_1) = \frac{1}{4}$. Again, using the location-scale relationship to $Z$, we first find the IQR of $Z$, then transform back to get the IQR of $X$. The conditions $$\Phi(q_1^*) = \frac{1}{4}, \quad \Phi(q_3^*) = \frac{3}{4}$$ are clearly symmetric (see part a). We can look up in a normal distribution table that $\Phi(-0.67449) \approx 0.25$, or to more precision with a computer, $$q_1^* \approx -0.67448975019608174320.$$ It follows that the IQR of $Z$ is $$q_3^* - q_1^* \approx 1.3489795003921634864,$$ hence the IQR of $X$ is $$q_3 - q_1 = (\sigma q_3^* + \mu) - (\sigma q_1^* + \mu) \approx 1.3489795 \sigma,$$ and so the desired ratio is simply approximately $$0.74130110925280093027.$$ Note this quantity does not depend on the parameters. Your error is that you performed the subtraction incorrectly.
Best Answer
Standard deviation and mean give a probability mass distribution, which is continuous. That means that you can't talk about discrete things (like saying "there are exactly $x$ elements greater than some cutoff $n$) just on the basis of mean and standard deviation. As such, you can't say with certainty that "there are no elements less than the lower bound $a$ or greater than the upper bound $b$", so you can't give discrete values to the bound $[a,b]$ over which your range would be defined.
What the mean and standard deviation of a normal distribution does let you do is make claims about the relative probability of an element existing at any spot along the distribution, so you could pick some set cutoff deviation (say, perhaps $3 \sigma$) and say that $99.7\%$ of the probability mass lies within a $\pm3\sigma$ bound.