Is it possible that two density functions that have the same mean and variance, but different distributions? Is it possible that two density functions that have the same mean and variance, but different distributions?
Can you give an example?
 A: Yes, of course. Consider a uniform distribution $\mathcal U[0,a]$, it has a mean $m$ and variance $v^2$ (excercise - calculate them). Now take a gaussian law $\mathcal N_{m,v^2}$ - it has the same mean and variance yet it defines a different distribution.
A: Besides actual examples, keep in mind that mean and variance are only two numbers which are used to give a (very rough) estimate of the distribution. This means that you have thrown away a lot of information (otherwise you would not talk about distributions, but just about the pair (mean, variance), so it's just natural that there are many distributions with those same values.
A: The smallest examples (i.e., smallest support for the probability distributions) can be obtained as follows. Choose a number $p$ with $0<p<1$. Let the first distribution give probability $p$ to the point $1/p$ and probability $1-p$ to the point $(-1)/(1-p)$.  Let the second distribution be the reflection of the first, i.e., give probability $p$ to the point $-1/p$ and probability $1-p$ to the point $1/(1-p)$. Both distributions have mean $0$ and variance $1/(p-p^2)$.
For a slightly less trivial example, now using three-point supports, choose any real number $a>1$ and consider the distribution giving the points $-a,0,a$ probabilities $p,1-2p,p$ where $p=1/2a^2$. This has mean $1$ and variance $1$ for all $a$.
A: Let $X$ and $Y$ be two random variables whose means $\mu_X$ and $\mu_Y$ and variances $\sigma^2_X$ and $\sigma^2_Y$  exist, and whose variances are non-zero.
We construct two close relatives $X^\ast$ and $Y^\ast$ of $X$ and $Y$ respectively such that $X$ and $Y$ have the same mean and variance.
First we do a shift to move the means to $0$. So let $X'=X-\mu_X$ and $Y'=Y-\mu_Y$. Then scale: let $X^\ast=\frac{X'}{\sigma_X}$ and $Y^\ast=\frac{Y'}{\sigma_Y}$.
The random variables $X^\ast$ and $Y^\ast$ each have mean $0$ and variance $1$.
With this method, we can take two random variables $X$ and $Y$ with very unlike distributions, like an exponential and a normal, or a normal and a discrete uniform, and by minor manipulation (shift, scale) transform them into random variables with the sa,me mean and variance. 
A: yes. suppose P(1-a)=1/4, P(1+a)=1/4; P(1-b)=1/4, P(1-b)=1/4. then it is obvious that the mean value is 1. The variance is 1/2(a^2+b^2). So just take a and b on a cycle.
A: This may help your intuition, if you know some high-school physics. Mean and variance are roughly analogous to centroid and moment of inertia of physical objects. And certainly you can have two different objects with the same centroid and moments of inertia, right?
