It's mathematically impossible to deduce mean or standard deviation from median/quartiles, because medians and quartiles discard most of the data on which the mean and standard deviation are based.
Example:
data frequency
0 50
1.4 4
2 50
That has a mean of 1.0 and standard deviation of 0.9. (I'm using 2 significant figures so I don't have to go into population versus sample standard deviation.)
data frequency
0 30
1.4 44
2 30
That data also has the median and quartiles the same as in your example, but now the mean is 1.2 and the standard deviation is 0.8.
data frequency
0 30
1.4 3
2 70
10000000 1
Now I've changed my maximum without changing the median or quartiles, you can see even more clearly how the median and quartiles exclude extreme data, because the mean is now 96000 and the standard deviation is 98000 (still 2 sig.fig.).
(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
Look at a table of the standard normal distribution.
At what values of z it is $\Phi(z)=0.25, \Phi(z)=0.50$ or $\Phi(z)=0.75$ ?
The values of z are the quartiles $Q_1,Q_2$ and $Q_3$.