Find cartesian equations.
For 1, I have $x=t(t-1), y=t(t-1)^2$, then $y/x= t-1$ and $t=(y/x)+1$. $x=(y^2+xy)/x^2$, then $x=(y^2+xy)^{1/3}$. Is there an explicit solution?
For 2, I have $t= \sin^{-1}(1-x)$ so $$y=\cos(\sin^{-1}(1-x))(1-\sin(\sin^{-1}(1-x)))= x^{3/2}(2-x)^{1/2}.$$ Is this correct ?
One could make the resolution of the second question a little less maddening by writing
$$\sin t \ = \ 1 - x \ \ \Rightarrow \ \ \cos t \ = \ \sqrt{1 - ( 1 - x )^2} \ = \ \sqrt{2x - x^2} $$
$$\Rightarrow \ \ y \ = \ \sqrt{2x - x^2} \ \cdot \ x \ , $$
etc. Your result does look to be correct.
Obviously, from the quadratic equation in t, we have $t=\dfrac{1\pm\sqrt{1+4x}}2$ : Plug this into $\dfrac yx=t-1$ to get an explicit solution for $y=f(x)$. The second point is also correct.
For the explicit solution to the first one you could use the quadratic formula:
$$y^2 + xy - x^3 = 0 \to y = \frac{-x \pm \sqrt{x^2 + 4 x^3}}{2} = x \frac{-1 \pm \sqrt{1 + 4x}}{2}.$$
