By the formulae in [Barth-Peters-Van de Ven, Chapter V] one has, for a surface which is complete intersection of type $(d_1, \ldots, d_{n-2})$ in $\mathbb{P}^n$:
$$c_1^2(X)= \big(\sum d_i-(n+1)\big)^2 \prod d_i,$$
$$c_2(X)=\bigg[\binom{n+1}{2}-(n+1)\sum d_i+\sum d_i^2 +\sum_{i \neq j} d_id_j \bigg]\prod d_i.$$

If I made correctly the computations (please check!), one obtains that the inequality
$$c_1^2(X) \leq 2 c_2(X)$$
is equivalent to
$$n+1 \leq \big(\sum d_i\big)^2,$$
and this is of course almost always true, since the right-hand term is $\geq (2(n-2))^2$.
So the answer to $1.$ seems to be *yes*.

For your question in the last comment, instead, the answer is clearly *no* if $S$ is ACM. In fact, every smooth surface $S$ with $H^1(S, \mathcal{O}_S)=0$ is ACM for some embedding in the projective space. Now take for instance a fake projective plane. It satisfies $p_g(S)=q(S)=0$, so it is ACM, but $$c_1^2(S)=3c_2(S).$$