Answer: \(\boxed{36}\), \(\boxed{0.0204}\), and \(\boxed{0.713}\).

For one throw of two fair dice, the probability of getting a pair of sixes is

\[p=\frac16\cdot\frac16=\frac{1}{36}.\]

So \(X\) has a geometric distribution with success probability \(p=\frac{1}{36}\).

(i) For a geometric distribution, the mean number of trials until the first success is

\[E(X)=\frac1p.\]

Therefore

\[E(X)=\frac{1}{1/36}=\boxed{36}.\]

(ii) Exactly 12 throws are required if the first 11 throws are not pairs of sixes and the 12th throw is a pair of sixes.

Thus

\[P(X=12)=\left(1-\frac{1}{36}\right)^{11}\left(\frac{1}{36}\right).\]

So

\[P(X=12)=\left(\frac{35}{36}\right)^{11}\frac{1}{36}=0.0204\quad\text{to 3 significant figures}.\]

(iii) More than 12 throws are required if the first 12 throws are all not pairs of sixes.

Therefore

\[P(X>12)=\left(1-\frac{1}{36}\right)^{12}.\]

So

\[P(X>12)=\left(\frac{35}{36}\right)^{12}=0.713\quad\text{to 3 significant figures}.\]