For a discrete probability distribution, the sum of all probabilities must equal 1.

For $x=1,2,3$: $P(X=x) = Cx^2$
$P(X=1) = C(1)^2 = C$
$P(X=2) = C(2)^2 = 4C$
$P(X=3) = C(3)^2 = 9C$

For $x=4,5,6$: $P(X=x) = C(7-x)^2$
$P(X=4) = C(7-4)^2 = C(3)^2 = 9C$
$P(X=5) = C(7-5)^2 = C(2)^2 = 4C$
$P(X=6) = C(7-6)^2 = C(1)^2 = C$

The constant C is $\boxed{\frac{1}{30}}$.

Substitute $C = \frac{1}{30}$ into the probabilities:
$P(X=1) = \frac{1}{30}$
$P(X=2) = \frac{4}{30}$
$P(X=3) = \frac{9}{30}$
$P(X=4) = \frac{9}{30}$
$P(X=5) = \frac{4}{30}$
$P(X=6) = \frac{1}{30}$

The mean of X is $\boxed{\frac{49}{15}}$.

First, calculate $E(X^2)$.

The variance of X is $\boxed{\frac{434}{225}}$.