Usually not. Denote $f_i=\mathbb{1}_{i\in S}$, then $\mathbb{E} f_i=\mathbb{E} f_i^2=kp_i$, $\sum f_i\equiv k$, $\mathbb{E} f_if_j=cp_ip_j$. Thus $$k^2p_i=k\mathbb{E} f_i=\mathbb{E} \sum_j f_if_j=cp_i(1-p_i)+kp_i.$$ If $p_i>0$, this gives $c(1-p_i)=k^2-k$, and if $p_i\ne p_j$ and $k>1$ this gives a contradiction.

Usually not. Denote $f_i=\mathbb{1}_{i\in S}$, then $\mathbb{E} f_i=\mathbb{E} f_i^2=kp_i$, $\sum f_i\equiv k$, $\mathbb{E} f_if_j=cp_ip_j$ if $i\ne j$. Thus $$k^2p_i=k\mathbb{E} f_i=\mathbb{E} \sum_j f_if_j=cp_i(1-p_i)+kp_i.$$ If $p_i>0$, this gives $c(1-p_i)=k^2-k$, and if $p_i\ne p_j$ and $k>1$ this gives a contradiction.