The second Friedrichs' inequalities?

The answer is no.

A pretty nice counter-example has been given by Stephen in this question: Friedrichs's inequality?

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Backstory 1: $\mathbf{H}_0(\operatorname{div};\Omega) \cap \mathbf{H}(\mathbf{curl};\Omega)$ and $\mathbf{H}(\operatorname{div};\Omega) \cap \mathbf{H}_{0}(\mathbf{curl};\Omega)$ are compactly embedded in $\mathbf{L}^2(\Omega)$. However, if the homogeneous boundary condition is only on part of the boundary, this compact embedding result is not true as of my knowledge. Rellich–Kondrachov compactness argument is used to prove the Poincaré-Friedrichs type inequality.

For the estimate, please refer to Vector potentials in three-dimensional non-smooth domains by Cherif Amrouche, Christine Bernardi, Monique Dauge and Vivette Girault.

Backstory 2: The function satisfies the boundary condition can be constructed implicitly using the following problem: $$ \left\{ \begin{aligned} -\Delta w = 0 & \text{ in }\Omega, \\ w = 1 & \text{ on }\Gamma_2, \\ \frac{\partial w}{\partial n} = 0& \text{ on }\Gamma_1. \end{aligned}\right. $$ Let the $\mathbf{u} = \nabla w$, we can see that this vector potential satisfies: $$ \nabla\times \mathbf{u} = 0 \text{ and } \nabla\cdot \mathbf{u} = 0, \\ \mathbf{n}\times \mathbf{u}\big|_{\Gamma_2} = 0,\text{ and } \mathbf{n}\cdot \mathbf{u}\big|_{\Gamma_1} = 0. $$ Yet $\mathbf{u}$ itself is totally non-trivial.