Mass of oxygen in He+O2 mixture
I don't see any flaws in your argumentation. I agree that $R, T, V = \mathrm{const}$, hence

$$\frac{p(\ce{He})}{n(\ce{He})} = \frac{p(\ce{O2})}{n(\ce{O2})}$$

$$m(\ce{O2}) = \frac{p(\ce{O2}) \cdot M(\ce{O2})}{p(\ce{He}) \cdot M(\ce{He}) } \cdot m(\ce{He})$$

Just one minor thing: you incorrectly converted partial pressure of helium: $\pu{42.5 torr}$ is $\pu{0.056 atm}$, not $\pu{0.559 atm}$, therefore your resulting mass should be 10 times higher. 

In fact, there was no need to convert $\pu{torr}$ to $\pu{atm}$ in the first place. Don't do unnecessary steps if you can avoid them. At the same time don't drop any units and always remember to include them in your final calculation.

$$m(\ce{O2}) = \frac{\pu{158 torr} \cdot \pu{31.99 g mol-1}}{\pu{42.5 torr} \cdot \pu{4.00 g mol-1}} \cdot \pu{1.42 g} = \pu{42.22 g}$$