Given your assumptions, this flow satisfies bernoulli. From the continuity equation, you can see that velocity will vary as the cross sectional area varies. Assuming the vertical distance change is very minor, it can be dropped from the bernoulli equation. From the continuity equation, v2 is less than v1, so that p2 is greater than p1. For a given cross section which is perpendicular to the flow, the magnitude of the velocity will be the same, since there are no friction forces acting on the fluid.
First note that the lack of a net force does not imply zero flow velocity since the question points out that the flow is inviscid.
Since the pipe has a uniform cross-section, and since there can be a flow ($v\gt 0$) then we can have $Q=Av\gt 0$ meaning answer (I) is not necessarily (always) true.
And if $Q$ is a constant and given $A$ is constant, then $v$ must also be constant, since $$Q=A_1v_1=A_2v_2=\text{constant} \\
\rightarrow A_1=A_2 \therefore v_1=v_2=\text{constant}$$ So we would chose (II).
For (III) to be true, then:
From Bernoulli's equation $$P_1 + \frac{1}{2} \rho v_{1}^2 + \rho gy_1 = P_2 + \frac{1}{2} \rho v_{2}^2 + \rho gy_2$$ we are told that the pressure is the same throughout the pipe, so the terms $P_1$ and $P_2$ are equal and so vanish, meaning $$\rho g \Delta y = \frac{1}{2}\rho (v_2^2-v_1^2)$$ For $v_1\ne v_2$ then $\Delta y\ne 0$ so that one end of the pipe is higher, but we established that $v_1=v_2$ so this means $\rho g\Delta y=0$ meaning the pipe is horizontal. This means that (III) can also be true.
To answer the other part of your question, we do not require that the continuity equation apply only for incompressible flows. The continuity equation applies to all fluids, whether they be compressible or incompressible flow because it expresses the law of conservation of mass which must be satisfied at every point in a flow.
The Bernoulli's equation $$P+\frac{1}{2}\rho v^{2}+\rho gh=\text{constant}$$ does require incompressibility though.
Also, the equation$^1$ you quote $$Q= \frac{dP}{R}$$ does not appear to be correct assuming that $P$ is pressure and $R$ is a length. The quantity on the left $Q$ has units $\frac{m^3}{s}$ while the RHS appears to have units $\frac{N}{m^3}$ and so does this not appear to be a valid equation.
$^1$If you are using a variation of the Hagen-Poiseuille equation, $$P=\frac{8\mu LQ}{\pi R^4}$$ where $$R=\frac{8\mu L}{\pi r^4}$$ is a measure of flow resistance, then it is valid.
Best Answer
Bernoulli's equation:
No the pressure won't be the same at all points in the pipe. Considering Pascal's Law, a change in pressure at any point in an enclosed fluid at rest is transmitted undiminished to all points in the fluid. But it holds for static fluids essentially. And equation of continuity stems from the principle of conservation of mass but in this case the mass is not conserved per unit time. Consider 2 different cross-sections of the pipe and fluid flow through it per unit time is not the same since some component of $g$ along with some resultant force is acting on the water column at the 2 different cross-sections and the 2 different cross-sections have a distinct height difference.So equation of continuity is not VALID in this case and hence conclusions cannot be made from this.
Use Bernoulli's principle and you get the relation $$P_2-P_1=(h_2-h_1)\rho g$$
Use the components of pressure in that direction and use the above equation.