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.

ADDED. Actually, this is also written in the book by Barth-Peters- Van de Ven. In Chapter V, at the beginning of the Section "The Geography of Chern Numbers", they say:

" The simplest examples, like complete intersections and double coverings of $\mathbb{P}^2$, pratically always yield a point of $D_1$ [where $D_1$ is the region in the $(c_1, c_2)$-plane given by $c_1^2 \leq 2c_2$]. Indeed, for a long time only few examples were known of surfaces with Chern pairs $(c_1^2, c_2)$ in $D_2$ [i.e., such that $2c_2 < c_1^2 \leq 3c_2$]."

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 $$c_1^2(S)=3c_2(S).$$

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 $\mathbb{P}^n$: $$c_1^2(X)= \big(\sum d_i-(n+1)\big)^2 \prod d_i,$$ $$c_2(X)=\bigg[\binom{n+1}{2}-(n+1)\sum d_i+\sum d_i^2 +\sum_{i < j} d_id_j \bigg]\prod d_i.$$

Then one obtains that the inequality $$c_1^2(X) \leq 2 c_2(X)$$ is equivalent to $$n+1 \leq \sum d_i^2,$$ and this is of course almost always true, since the right-hand term is $\geq 4(n-2)$. So the answer to $1.$ seems to be yes.

ADDED. Actually, this is also written in the book by Barth-Peters- Van de Ven. In Chapter V, at the beginning of the Section "The Geography of Chern Numbers", they say:

" The simplest examples, like complete intersections and double coverings of $\mathbb{P}^2$, pratically always yield a point of $D_1$ [where $D_1$ is the region in the $(c_1, c_2)$-plane given by $c_1^2 \leq 2c_2$]. Indeed, for a long time only few examples were known of surfaces with Chern pairs $(c_1^2, c_2)$ in $D_2$ [i.e., such that $2c_2 < c_1^2 \leq 3c_2$]."

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 $$c_1^2(S)=3c_2(S).$$

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 $$c_1^2(S)=3c_2(S).$$

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.

ADDED. Actually, this is also written in the book by Barth-Peters- Van de Ven. In Chapter V, at the beginning of the Section "The Geography of Chern Numbers", they say:

" The simplest examples, like complete intersections and double coverings of $\mathbb{P}^2$, pratically always yield a point of $D_1$ [where $D_1$ is the region in the $(c_1, c_2)$-plane given by $c_1^2 \leq 2c_2$]. Indeed, for a long time only few examples were known of surfaces with Chern pairs $(c_1^2, c_2)$ in $D_2$ [i.e., such that $2c_2 < c_1^2 \leq 3c_2$]."

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 $$c_1^2(S)=3c_2(S).$$