 Solution

All three assertions are false. For the transversality assertion, use $f:\mathbb{R}\to \mathbb{R}$, $f(x)=x^3$, and $Z=\{0\}$. At the preimage point the derivative is zero, so $f$ is not transverse to $Z$, yet $f^{-1}(Z)=\{0\}$ is a codimension-one submanifold. Transversality is sufficient, not necessary.

For openness of regular values, the smooth function $f(x)=e^{-x}(2+3\sin x)$ gives a connected-domain counterexample. Every zero has $\sin x=-2/3$ and derivative $3e^{-x}\cos x\ne 0$, so zero is a regular value. Critical points solve $\cos x-\sin x=2/3$. Choose any such point $x_0$; its value is nonzero, since simultaneous vanishing would force $\sin x=-2/3$ and $\cos x=0$. The critical points $x_0+2\pi n$ have nonzero critical values $e^{-2\pi n}f(x_0)\to 0$. Thus no neighborhood of the regular value zero consists entirely of regular values. Properness would supply the missing compactness needed for the usual open-regular-value conclusion.

For the measure assertion, a constant map $f:\mathbb{R}\to \mathbb{R}$ has every point critical. Its critical-point set has positive, indeed infinite, measure. Sard theorem concerns the measure of the set of critical values in the target, not the measure of critical points in the domain.