Minimum of variance The probability that $X \in (-1, 3)$ equals $0$ ;
Expectation: $\mathbb{E} X=0$
How to find minimal variance?
I suppose that the answer is 3. Is that right? And how to prove it?
 A: Hint: the smallest variance will occur when the entire probability mass is at $−1$ and $3$ and distributed so that the mean is $0$.
A: We should use Markov's inequality.Note that 
$$P(-2<(X-1)<2)=0$$
Consequently, 
$$P((X-1)^2<4)=0$$ 
Next, Markov inequality for $Y=(X-1)^2$:
$$P(Y \geq 4) \leq E(Y)/4 = E(X^2-2X+1)/4 = (E(X^2)+1)/4$$
Because $E(X) = 0$. Then, $D(X)=E(X^2)$. So, 
$$E(X^2)\geq P(Y \geq 4)-1=3$$
A: Let $X$ be a random variable such that $\mathbb{E}X=0$, $\mathbb{P}[X \in (-1,3)]=0$. You want to solve
$$\mathbb{E}((X-\mathbb{E}X)^2)=\mathbb{E}(X^2) = \int_{[X \leq -1]} X^2 \, d\mathbb{P}+ \int_{[X \geq 3]} X^2 \, d\mathbb{P} \to \min$$
One can easily see that the minimum is attained for $X \sim p_1 \cdot \delta_{-1}+p_2 \cdot \delta_{3}$. From the condition $\mathbb{E}X=0$ and $p_1+p_2=1$ you obtain $p_1$, $p_2$. 
A: $\newcommand{\var}{\operatorname{var}}\newcommand{\E}{\operatorname E}$
Let $$X=\begin{cases} 3 & \text{with probability }1/4, \\ -1 & \text{with probability }3/4. \end{cases}$$
This puts the values that $X$ can assume as close to $0$ as they can get without violating the constraints you've set.  We get $\E(X)=0$ and $\var(X) = 3$.

Later edit: Suppose $Y$ is some other r.v. satisfying the constraints.  Then
$$
\E(Y) = \E(Y\mid Y\ge3)\Pr(Y\ge3)+\E(Y\mid Y\le-1)\Pr(Y\le-1)=0,
$$
and
\begin{align}
\var(Y) = {} & \E(\var(Y\mid 1_{Y\ge3}))+\var(E(Y\mid 1_{Y\ge3})) \\[12pt]
= {} & \overbrace{\var(Y\mid Y\ge3)\Pr(Y\ge3)+\var(Y\mid Y\le-1)\Pr(Y\le-1)}^\text{unexplained component of the variance} \\[4pt]
& {}+ \overbrace{\underbrace{(\E(Y\mid Y\ge3)-\E(Y\mid Y\le-1))^2}_\text{square}\  \underbrace{\Pr(Y\ge3)\Pr(Y\le-1)}_\text{product}}^\text{explained component of the variance}.
\end{align}
The "unexplained" component of the variance will be $0$ precisely in the case ofthe r.v. $X$ defined above.
The "product" is of the form $p(1-p)$ and gets bigger as $p$ gets closer to $1/2$ and smaller as $p$ gets closer to either endpoint.  The "square" is minimized subect to the required constraints by the variable $X$ defined above.  So our only hope of making $\var(Y)$ smaller than $3=\var(X)$ must lie in making the "product" small.  That requires making $p$ closer to $1$ or to $0$.
Probably I'll continue this further later . . . . . . .
