Looks like the third one is true as well

\\((\hat{F} \circ \hat{G})(x, z)\\)

\\(= \bigvee\big(\hat{G}(x, y) \otimes \hat{F}(y, z)\big)\\)

\\(= \bigvee\big(Y(Gx, y) \otimes Z(Fy, z)\big)\\)

And it turns out this join \\(= Z(FGx, z) = \widehat{FG}(x, z)\\)

To prove this, first set \\(y = Gx\\) to see that our join

\\(\geq Y(Gx, Gx) \otimes Z(FGx, z) \geq I \otimes Z(FGx, z) = Z(FGx, z)\\)

But given any \\(y\\) we have

\\(Z(FGx, z) \geq Z(FGx, Fy) \otimes Z(Fy, z) \geq Y(Gx, y) \otimes Z(Fy, z)\\)

So we also have \\(Z(FGx, z) \geq\\) our join. QED