You're not wrong, here's a snipet of R that validates your results:
> x <- c(1, 2, 3, 3, 4, 4, 5, 5, 5, 5)
> y <- 24.75 * x - 23.75
> mean(x)
[1] 3.7
> mean(y)
[1] 67.825
> mean(x*x) - mean(x)**2
[1] 1.81
> mean(y*y) - mean(y)**2
[1] 1108.738
The essential point that you are missing is that the variance of a random variable has different units than the random variable itself, so you should not expect them to have the similar magnitudes. For example, human height has length like units - say we measure in meters - the variance of human height then has units meters^2. The standard deviation is defined in the way it is to recalibrate the measure of spread back into the same units as the random variable itself. This is why you never, for example, see variance bands on a histogram, it's always standard deviation/error bands, as these have the appropriate units of measurement.
Mean minimizes the squared error (or L2 norm, see here or here), so natural choice for variance to measure distance from the mean is to use squared error (see here on why we square it). On the other hand, median minimizes the absolute error (L1 norm), i.e. it is a value that is in the "middle" of your data, so absolute distance from the median (so called Median Absolute Deviation or MAD) seems to be a better measure of the degree of variability around the median. You can read more about this relations in this thread.
Saying it short, variance differs from MAD on how do they define the central point of your data and this influences the way how we measure variation of datapoints around it. Squaring the values make outliers have greater influence on the central point (mean), while in case of median, all the points have the same impact on it, so the absolute distance seems more appropriate.
This can be shown also by simple simulation. If you compare values squared distances from the mean and median, then the total squared distance is almost always smaller from mean than from median. On the other hand, total absolute distance is smaller from median, then from mean. The R code for conducting the simulation is posted below.
sqtest <- function(x) sum((x-mean(x))^2) < sum((x-median(x))^2)
abstest <- function(x) sum(abs(x-mean(x))) > sum(abs(x-median(x)))
mean(replicate(1000, sqtest(rnorm(1000))))
mean(replicate(1000, abstest(rnorm(1000))))
mean(replicate(1000, sqtest(rexp(1000))))
mean(replicate(1000, abstest(rexp(1000))))
mean(replicate(1000, sqtest(runif(1000))))
mean(replicate(1000, abstest(runif(1000))))
In the case of using median instead of mean in estimating such "variance" this would lead to higher estimates, than with using mean as it is done traditionally.
By the way, the relations of L1 and L2 norms can be considered also in the Bayesian context, as in this thread.
Best Answer
If you only have the sequence of moments, the $\frac12$-th moment is not necessarily determined by them.
If the MGF exists in a neighborhood of zero, then the moment sequence would determined the distribution and the $\frac12$-th moment should be determined (though not always amenable to algebraic calculation).
However if you have the pdf, we can avoid all that, since we can try to compute $E(X^\frac12)$ directly (e.g. by calculating the integral $\int_0^\infty x^\frac12 f(x) dx$ -- I presume $X$ is on the non-negative half line,for the obvious reason). Note also that the distribution of $\sqrt{X}$ will be very easy to write down (if $Y=\sqrt{X},\, F_Y(y)=F_X(y^2)$ and $f_Y(y)=2yf_X(y^2)$). If, for example, we recognize that density as a standard one - it might be very fast to identify the expectation that way.
As whuber points out in comments, all we need to find is $E(\sqrt{X})$, since $\text{Var}(\sqrt{X})=E(X)-E(\sqrt{X})^2$.