Q1. We can prove your answer to Q1 for small $N$ as follows.
Note first that we can replace each number $x$ to hit by $\lceil x\rceil$, nondecreasing the score. Moreover, no number $x$ bigger than $2^{N-1}$ can be hit. That is, we can suppose that the set $X_N$ of numbers to hit is $\{1,2,\dots, 2^{N-1}\}$ and each natural number $x\in X$ has a probability $p_x$ in the distribution $D$.
For any natural numbers $n\le N$ and $x\le 2^{N-1}$ let $E(n,x)$ be the expectation of the score when there remains $n$ numbers to hit and the current controlling number is $x$. Then $$E(1,x)=\sum_{y=1}^x (x+y)p_y\mbox{ and }E(n,x)=\sum_{y=1}^x E(n-1,x+y)p_y$$ for $n>1$. This easily implies that the required expectation $E(N,1)$ of the score is equal to$$\sum\{(x_0+\dots+x_N)p_{x_1}\dots p_{x_N}: x_0=1\mbox{ and }X\ni x_i\le\sum_{j=0}^{i-1} x_j\mbox{ for each natural }i\le N\}.$$
Now we can answer the question for small $N$.
Let $N=2$. Then in the sum for $E(N,1)$ we have the following sequences $(x_1,x_2)$: $(1,1)$ and $(1,2)$. Then$$E(N,1)=3p_1^2+4p_1p_2=p_1(3p_1+4(1-p_1))=p_1(4-p_1),$$that is maximized when $p_1=1$.
$N=3$Let $N=2$. Then in the sum for $E(N,1)$ we have the following sequences $(x_1,x_2,x_3)$ with the corresponding sums $\sum_{i=0}^N x_i$ and the products of probabilities:
$(1,1,1):$$4$$p_1p_1p_1$
$(1,1,2):$$5$$p_1p_1p_2$
$(1,1,3):$$6$$p_1p_1p_3$
$(1,2,1):$$5$$p_1p_2p_1$
$(1,2,2):$$6$$p_1p_2p_2$
$(1,2,3):$$7$$p_1p_2p_3$
$(1,2,4):$$8$$p_1p_2p_4$
Thus$$E(N,1)=p_1(4p_1^2+10p_1p_2+6p_1p_3+6p_2^2+7p_2p_3+8p_2p_4).$$
Replacing $p_4$ by $1-p_1-p_2-p_3$, we obtain
$$E(N,1)=p_1(4p_1^2+2p_1p_2+6p_1p_3-2p_2^2-p_2p_3+8p_2).$$
Increasing $p_1$, if needed, we can assume that $p_1+p_2+p_3=1$.Then replacing $p_3$ by $1-p_1-p_2$, we obtain
$$E(N,1)=p_1(-2p_1^2-3p_1p_2-p_2^2+6p_1+7p_2).$$
Since $\frac{\partial E(N,1)}{\partial p_2}=p_1(-3p_1-2p_2+7)\ge 0$,increasing $p_2$, if needed, we can assume that $p_1+p_2=1$.Then replacing $p_2$ by $1-p_1$, we obtain
$$E(N,1)=p_1(6-2p_1),$$that is maximized when $p_1=1$.
