Var(X) is known, how to calculate Var(1/X)? If I have only $\mathrm{Var}(X)$, how can I calculate $\mathrm{Var}(\frac{1}{X})$?
I do not have any information about the distribution of $X$, so I cannot use transformation, or any other methods which use the probability distribution of $X$. 
 A: You can use Taylor series to get an approximation of the low order moments of a transformed random variable. If the distribution is fairly 'tight' around the mean (in a particular sense), the approximation can be pretty good.
So for example
$$g(X) = g(\mu) + (X-\mu) g'(\mu) + \frac{(X-\mu)^2}{2} g''(\mu) + \ldots$$
so
\begin{eqnarray}
\text{Var}[g(X)] &=& \text{Var}[g(\mu) + (X-\mu) g'(\mu) + \frac{(X-\mu)^2}{2} g''(\mu) + \ldots]\\
 &=& \text{Var}[(X-\mu) g'(\mu) + \frac{(X-\mu)^2}{2} g''(\mu) + \ldots]\\
 &=& g'(\mu)^2 \text{Var}[(X-\mu)] + 2g'(\mu)\text{Cov}[(X-\mu),\frac{(X-\mu)^2}{2} g''(\mu) + \ldots]  \\& &\quad+ \text{Var}[\frac{(X-\mu)^2}{2} g''(\mu) + \ldots]\\
\end{eqnarray}
often only the first term is taken
$$\text{Var}[g(X)] \approx g'(\mu)^2 \text{Var}(X)$$
In this case (assuming I didn't make a mistake), with $g(X)=\frac{1}{X}$, $\text{Var}[\frac{1}{X}] \approx \frac{1}{\mu^4} \text{Var}(X)$.
Wikipedia: Taylor expansions for the moments of functions of random variables
---
Some examples to illustrate this. I'll generate two (gamma-distributed) samples in R, one with a 'not-so-tight' distribution about the mean and one a bit tighter.
 a <- rgamma(1000,10,1)  # mean and variance 10; the mean is not many sds from 0
 var(a)
[1] 10.20819  # reasonably close to the population variance

The approximation suggests the variance of $1/a$ should be close to $(1/10)^4 \times 10 = 0.001$
 var(1/a)
[1] 0.00147171

Algebraic calculation has that the actual population variance is $1/648 \approx 0.00154$
Now for the tighter one:
 a <- rgamma(1000,100,10) # should have mean 10 and variance 1
 var(a)
[1] 1.069147

The approximation suggests the variance of $1/a$ should be close to $(1/10)^4 \times 1 = 0.0001$
 var(1/a)
[1] 0.0001122586

Algebraic calculation shows that the population variance of the reciprocal is $\frac{10^2}{99^2\times 98} \approx 0.000104$.
A: It is impossible.
Consider a sequence $X_n$ of random variables, where 
$$P(X_n=n-1)=P(X_n=n+1)=0.5$$
Then: 
$$\newcommand{\Var}{\mathrm{Var}}\Var(X_n)=1 \quad \text{for all $n$}$$
But $\Var\left(\frac{1}{X_n}\right)$ approaches zero as $n$ goes to infinity:
$$\Var\left(\frac{1}{X_n}\right)=\left(0.5\left(\frac{1}{n+1}-\frac{1}{n-1}\right)\right)^2$$
This example uses the fact that $\Var(X)$ is invariant under translations of $X$, but $\Var\left(\frac{1}{X}\right)$ is not.
But even if we assume $\mathrm{E}(X)=0$, we can't compute $\Var\left(\frac{1}{X}\right)$:
Let 
$$P(X_n=-1)=P(X_n=1)=0.5\left(1-\frac{1}{n}\right)$$
and 
$$P(X_n=0)=\frac{1}{n} \quad \text{for $n>0$} $$
Then $\Var(X_n)$ approaches 1 as $n$ goes to infinity, but $\Var\left(\frac{1}{X_n}\right)=\infty$ for all $n$.
