At Normal and Binomial models, Is always the posterior variance less than the prior variance? Or what conditions guarantee that? 
In general (and not only normal and binomial models) I suppose the main reason that broke this claim is that there's inconsistency between the sampling model and the prior, but what else? 
I'm starting with this topic, so I really appreciate easy examples
 A: This is going to be more of a question to @Xi'an than an answer. 
I was going to answer that a posterior variance 
$$
V(\theta|y)=\frac{\alpha_{1}\beta_1}{(\alpha_{1}+\beta_1)^2(\alpha_{1}+\beta_1+1)}=\frac{(\alpha _{0}+k)(n-k+\beta_{0})}{(\alpha_{0}+n+\beta_0)^2(\alpha_{0}+n+\beta_0+1)},
$$
with $n$ the number of trials, $k$ the number of successes and $\alpha_{0},\beta_0$ the coefficients of the beta prior, exceeding the prior variance 
$$
V(\theta)=\frac{\alpha_{0}\beta_0}{(\alpha_{0}+\beta_0)^2(\alpha_{0}+\beta_0+1)}
$$
is possible also in the binomial model based on the example below, in which likelihood and prior are in stark contrast so that the posterior is "too far in between". It does seem to contradict the quote by Gelman.
n <- 10         
k <- 1
alpha0 <- 100
beta0 <- 20

theta <- seq(0.01,0.99,by=0.005)
likelihood <- theta^k*(1-theta)^(n-k) 
prior <- function(theta,alpha0,beta0) return(dbeta(theta,alpha0,beta0))
posterior <- dbeta(theta,alpha0+k,beta0+n-k)

plot(theta,likelihood,type="l",ylab="density",col="lightblue",lwd=2)

likelihood_scaled <- dbeta(theta,k+1,n-k+1)
plot(theta,likelihood_scaled,type="l",ylim=c(0,max(c(likelihood_scaled,posterior,prior(theta,alpha0,beta0)))),ylab="density",col="lightblue",lwd=2)
lines(theta,prior(theta,alpha0,beta0),lty=2,col="gold",lwd=2)
lines(theta,posterior,lty=3,col="darkgreen",lwd=2)
legend("top",c("Likelihood","Prior","Posterior"),lty=c(1,2,3),lwd=2,col=c("lightblue","gold","darkgreen"))

 > (postvariance <- (alpha0+k)*(n-k+beta0)/((alpha0+n+beta0)^2*(alpha0+n+beta0+1)))
[1] 0.001323005
> (priorvariance <- (alpha0*beta0)/((alpha0+beta0)^2*(alpha0+beta0+1)))
[1] 0.001147842

Hence, this example suggests a larger posterior variance in the binomial model.
Of course, this is not the expected posterior variance. Is that where the discrepancy lies?
The corresponding figure is
 
A: Since the posterior and prior variances on $\theta$ satisfy (with $X$ denoting the sample)
$$\text{var}(\theta) = \mathbb{E}[\text{var}(\theta|X)]+\text{var}(\mathbb{E}[\theta|X])$$ assuming all quantities exist, you can expect the posterior variance to be smaller on average (in $X$). This is in particular the case when the posterior variance is constant in $X$. But, as shown by the other answer, there may be realisations of the posterior variance that are larger, since the result only holds in expectation.
To quote from Andrew Gelman,

We consider this in chapter 2 in Bayesian Data Analysis, I think in a
  couple of the homework problems. The short answer is that, in
  expectation, the posterior variance decreases as you get more
  information, but, depending on the model, in particular cases the
  variance can increase. For some models such as the normal and
  binomial, the posterior variance can only decrease. But consider the t
  model with low degrees of freedom (which can be interpreted as a
  mixture of normals with common mean and different variances). if you
  observe an extreme value, that’s evidence that the variance is high,
  and indeed your posterior variance can go up.

