I think you need to make this question more stand alone. Foe example, 1 doesn't make much sense on its own. For 2, you need to understand some Clifford theory. An irreducible representation over $F_{p}$ of a cyclic group $\langle z \rangle$ of order $n$ becomes over the algebraic closure, a direct sum of distinct absolutely irreducible representations which are conjugate via the Frobenius automorphism $x \to x^{p}.$ So you need to look at the action of this automorphism on the $1$-dimensional representations of $\langle z \rangle.$ Since $z$ is conjugate to $z^{-1}$ within the dihedral group, any $F_{p}$ representation of the cyclic group which is to have any chance of extending to the dihedral group has to have $1$-dimensional representation and its dual in the same orbit under that Frobenius automorphism.
When you think about what this means, you get condition 2 ( strictly, we have only demonstrated necessity here, but sufficiency is similar).
For 3,4 Any faithful representation of the dihedral group has to "lie over" a faithful representation
of the cyclic subgroup of index $2.$ On the other hand, if it does lie over such a faithful representation, then since every element outside the cyclic subgroup has order $2$ and inverts every element of the cyclic subgroup, no element outside the cyclic subgroup can lie in the kernel of the representation, while no non-identity element of the cyclic subgroup is in the kernel, so only the identity is in the kernel.
