In Bayesian Statistic how do you usually find out what is the distribution of the unknown? To estimate the posterior we have
$$p(\theta|x) = \frac{p(\theta)*p(x|\theta)}{\sum p(\theta ')*p(x|\theta ')}$$
$x$ is usually the experimentally sampled data, and $\theta$ is the model, but both $p(x|\theta)$ and $p(\theta)$ is unknown, how do you usually measure those two quantities?
 A: These quantities are known as part of the model. 
$p(\theta)$ is the prior, which you chose (a classic example is the Beta distribution), and $p(x|\theta)$ is the density function of $X|\theta$, for example the model is such that $X|\theta\sim\mathcal{N}(\theta,1)$.
A: The prior distribution is based on previous experience, data, intuition, whatever. If none of these are available, then a flat or non-informative prior is given. 
Presumably, you know the distributional form of $p(x|\theta)$ based on your experiment.
If possible, it is desirable to pick a prior distribution $p(\theta)$ that is
"conjugate" to the likelihood (data) distribution $p(x|\theta)$.
That is, "mathematically compatible". In that case one can deduce
the posterior distribution $p(\theta|x)$ without having to compute the denominator.
Simple example:  Trying to predict the outcome of an election, a
campaign strategist chooses the prior distribution $Beta(\alpha_0 = 330, \beta_0 = 270)$ on the probability $\theta$ that the candidate
is going to win. This is based on a 'hunch' that $\theta$ is 'near'
0.55 and is most likely to lie somewhere in $(0.51, 0.59)$. [You can check that this beta distribution has about the right properties.]
A subsequent reliable poll shows $x = 620$ in $n = 1000$ favoring
the candidate. This gives a binomial likelihood. Ignoring the
constants that make the beta and binomial distributions sum to 1,
we have 
$$p(\theta|x) \propto p(\theta)p(x|\theta) 
\propto \theta^{\alpha_0 -1} (1-\theta)^{\beta_0 -1} \times
\theta^x (1-\theta)^{n-x}\\ = 
\theta^{a_0 + x - 1}(1-\theta)^{\beta_0 - n - x -1}
= \theta^{\alpha_n -1}(1-\theta)^{\beta_n - 1},$$
where we recognize the posterior is proportional to ($\propto)$
the density of the distribution $Beta(\alpha_n, \beta_n),$
with $\alpha_n = 950$ and $\beta_n = 650.$
Then one can cut 2.5% of the area from each tail of the posterior
distribution to find the posterior probability interval $(0.57,0.62).$ This is different from the frequentist 95% CI
one would have obtained from the data alone. One can say that
the Bayesian approach has accomplished an appropriate melding
of prior opinion and data to produce a probability interval
estimate for $\theta.$
In practice, one would typically try several different prior
distributions in order to assess the influence of each on the
result. (Very roughly speaking, our strategist's prior distribution carries
about the same weight as a poll of 600 people with 330 in
favor of our candidate.)
If the consultant is from Mars and has no knowledge of human
elections, then the non-informative prior distribution might
have been something like $Beta(1,1)$ or $Beta(.5,.5)$ and
the endpoints of the posterior probability interval would closely approximate the endpoints of the frequentist CI based on the data
alone. (The philosophical interpretations of the Bayesian and
frequenist intervals are rather different, but that is for another
discussion.)
Another Answer suggests the prior might be beta and the likelihood normal.
Computationally speaking, that would be a bit messier because
beta and normal are not conjugate distributions and we would need
to compute the denominator of Bayes' theorem. Beta is conjugate to
binomial; normal is conjugate to normal (but not quite obviously so).
It is worth noting that graphs of densities of $Norm(\mu=.55,\sigma=.02)$ and $Beta(330, 270)$ are difficult
to distinguish [except, or course, the normal is not constrained
to $(0,1).$]
Acknowledgment: This example is similar to one in Chapter 8 of
Suess and Trumbo (2010), Springer.
