Concur with this point:
"$\ce{2s}$ and $\ce{2p}$ valence electrons" is not the same thing as "two $\ce{s}$ and two $\ce{p}$ valence electrons".
Your book probably used the latter, but you interpreted it as the former. Though, it is entirely possible that the book wrote it the way you say it had, but unless you quote that particular line and cite the book (as Mod. Martin pointed out), it will be impossible to ascertain that.
So for the record, silicon does have a valence configuration of $$\ce{3s^{2} 3p^{2}}$$
You must be attentive while reading such things ;-)
Now for the second part of your question,
Then it says the maximum number of electrons in the outermost shell is $\ce{8}$
Well, technically, the maximum possible number of electrons in the outermost shell (which is $\ce{n = 3}$ in this case) is $\ce{18}$ and not $\ce{8}$.
That's because the third shell is constituted by $\ce{s, p, d}$ subshells. (Granted, ground-state silicon doesn't use those $\ce{d}$ orbitals, but your post only mentions "outermost shell", so I thought mentioning this should be of help)
The third shell would therefore have one $\ce{s}$ orbital (holds 2 electrons), three $\ce{p}$ orbitals (holds 6 electrons) and five $\ce{d}$ orbitals (holds 10 electrons). So,
$$\mathrm{ 2 + 6 + 10 = 18 }$$