There are a number of questions involved in this post so even if you do not feel able to answer all please answer any that you can. Consider a stochastic process $Z(x)$ where $x$ is the index parameter. This process is said to be mean-square continuous if $\mathbb{E}\left[Z(x)^2\right] < +\infty$ and:
\begin{equation} \lim\limits_{x \rightarrow s} \mathbb{E}\left[\left(Z(x) - Z(s)\right)^2\right] = 0 \end{equation}
Question 1): Are these two conditions necessary and sufficient or are other conditions required for mean square continuity of $Z(x)$?
Question 2): Does the differentiability of realisation $z(x)$, also known as a sample path, of $Z(x)$ imply the existence of a derivative process $\dot{Z}(x)$ with finite variance?
Question 3): Is saying that $Z(x)$ is differentiable nowhere equivalent to saying that its derivative process $\dot{Z}(x)$ has infinite variance? For instance the Wiener process is mean-square continuous yet nowhere differentiable and is often used to represent the integral of white noise which has infinite variance.
Question 4) Is it correct to say that mean-square differentiability of $Z(x)$ is a necessary but not sufficient condition for the differentiability of any realisation $z(x)$? But rather that if a process is both mean-square differentiable and separable then its samples paths are differentiable.