If there are $a$ dupvotes, $b$ upvotes, and $c$ downvotes, then this problem is asking for how many ordered triplets of $(a,b,c)$ satisfy the following system of equations:

$${\displaystyle 2a+b-c=P}$$ $${\displaystyle a+b+c=U}$$ $${\displaystyle x:y=R1:R2}$$

where $x$ and $y$ could be any two of $a$, $b$, and $c$.

No need to do anything fancy for the first problem. Just brute force all possibilities of upvotes, downvotes, and dupvotes and see how many of them add to the net points and satisfy the ratio. Don't be fooled! Although there are three unknowns, the solution is not $\mathcal{O}(N^3)$. You only need to brute force $a$ and $b$. Since $U$ is given, $c$ is already solved as $U-a-b$. Another thing to note is that the statement mentioned that all three values will be positive, so we need not worry about infinite ratios.

Real numbers are icky, and reducing the ratios is just too much work. To check if two ratios $x:y$ and $R1:R2$ are equal, we can simply cross multiply and see if $x\times R2$ equals $y\times R1$.