Understanding the Bayesian question I have a Bayesian question here:

In a drug experiment, patients with a chronic condition are asked to choose between two drugs, C (control), and T (new treatment). (You may assume for the purpose of this question that every patient will express a preference one way or the other). Let the population proportion who prefer “T” be θ. We observe a sample of n patients which their responses are independent given θ. 
Elicit three prior distributions for θ as follows: 
  1) $π_1 (θ)$, a non-informative (or vague) prior; 
2) $π_2 (θ)$, a proper informative prior where its mean and standard deviation are 0.5 and  0.25, respectively; 
3) $π_3 (θ)$,  a proper informative prior where its first and third quartiles, are Q1=0.62, and Q3=0.715, respectively, and its median is 0.67. 

I have a problem on understanding the question above. Is it correct if I define the population proportion as $π(θ|T)$?
I really don't think I understand the relationship between T and θ probably, I would be grateful if someone who can explain it to me. 
Although I don't get the question, I have answered 1) and 3), but not sure about 2).
For 1), I defined it as Be(1,1) ( from θ~U[0,1], all data are equally likely so it is non-informative)
For 3), I used the elicitation tool(http://optics.eee.nottingham.ac.uk/match/uncertainty.php#) to get Be(29,14) as answer.
For 2), I have tried to apply $$E(X)=\frac{x}{x+B}=0.5$$ and $$Var(X)=\frac{xB}{(x+B)^2(x+B+1)}=0.25^2$$
But it does not work as I get x = B from E(X).
Is there any other way I can use to solve this, either by hand or using software?
Any help would be appreciated.
 A: (1) One possible non-informative prior for $\theta$ would be $\theta \sim \mathsf{Beta}(\alpha =1, \beta=1),$ as you say.  Based on sound theoretical arguments,
some Bayesian statisticians would prefer the Jeffreys prior
$\mathsf{Beta}(\alpha =1/2, \beta=1/2).$ [If it's not in your text, you could google 'Jeffreys prior'.]
(2) If you want a beta prior with mean $\mu = 1/2$ then you need the shape
parameters to be equal: $\alpha = \beta.$ Next, you can find the value of $\alpha$ that makes the variance $\sigma^2 = \frac{\alpha^2}{4\alpha^2(2\alpha+1)}=1/16.$ Solving for $\alpha (= \beta)$ seems to give a reasonable answer.
(3) You have chosen $\alpha = 29,\, \beta = 14$. Then the respective quartiles are about
0.62, 0.67, 0.72, which are very nearly your target values. [The shape parameters of the beta distribution need not be integers, so you might be able to get a little closer to the target values.]
qbeta(c(.25, .5, .75), 28, 14)
[1] 0.6189876 0.6693351 0.7172090

Addendum on notation: There is no way for me to know the notation of your text, and the notation you are using does not make sense. Perhaps this will help: 
The success probability $\theta$ is viewed as a random variable. Its beta prior distribution has density function proportional to
$\pi(\theta) \propto \theta^{1 - \alpha}(1 - \theta)^{\beta - 1}.$ 
If your data shows $x$ successes in $n$ trials, then the likelihood function is proportional to $\pi(x|\theta) \propto \theta^x(1-\theta)^{n-x}.$ [Notation $\propto$ is read 'proportional to'.]
Finally, by Bayes' Theorem, the posterior distribution is proportional to $\pi(\theta | x) \propto 
\theta^{\alpha + x - 1}(1-\theta)^{\beta + n - x-1}.$ This is the product of the prior and the likelihood. 
Because the beta prior is 'conjugate' to the binomial likelihood, we can recognize the form of the posterior distribution $\pi(\theta|x)$ as the kernel of $\mathsf{Beta}(\alpha + x, \beta + n - x).$ [A 'kernel' of a density function is the density function without the norming constant that makes it integrate to $1.$ For example, the density function of a beta random variable is $\frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta)}\theta^{\alpha-1}(1-\theta)^{\beta-1}$
for $0 < \theta < 1.$ Its kernel is $\theta^{\alpha-1}(1-\theta)^{\beta-1}.]$
