{\partial(x,y)\over\partial(z,w)}=\begin{bmatrix}w&z\\0&1\end{bmatrix} {f_Z(z)}=\int_{0}^\infty f_X(x)f_Y(y)\Big{|}{\partial(x,y)\over\partial(z,w)}\Big{|}dw=\lambda^2\int_{0}^\infty we^{-(z+1)\lambda w}dw={1\over(z+1)^2}

Hint: Let (x,y) \in C^2, \tau \in [0,1]. We want to show that x + \tau(y-x) \in C. Let \epsilon > 0. Given that x + j\cdot 2^{-n}(y-x) \in C for all n and j \in \{0,\dots, 2^n\} ( why ...

You can try some post-processing after calculating of x: compute the null space of E, then solve \min \|x + v \| subject to Ev=0 and x+v\ge 0.

Now, use \sqrt{x(y+z)}-\sqrt{y(x+z)}=\frac{x(y+z)-y(x+z)}{\sqrt{x(y+z)}+\sqrt{y(x+z)}}=\frac{z(x-y)}{\sqrt{x(y+z)}+\sqrt{y(x+z)}}.

In practice, the apparatus measuring the spin should be localized somewhere in space (it cannot fill the whole universe!) and this fact implies that you always make a measurement of position (actually ...

You just mixed up some letters. xyz=1 implies yzx=1 and zxy=1, not yxz=1.