I will only hint at the solution, leaving it to you to make things explicit. This way you will have to chase through the definitions, which, I claim, is a good exercise.
For simplicity I assume this is about 2-dimensional surfaces in $\mathbb{R}^3$. I also assume you know the 2nd fundamental form of a hypersurface can be expressed as the (covariant) derivative of the normal to the hypersurface along that hypersurface (up to sign).
If two surfaces $M_1, M_2$ touch along a curve, they share the normal along the curve (up to sign) (why?). The curve is a line of curvature iff it's tangent is an eigenvector of the 2nd fundamental form. This means: if $\phi$ parametrizes the curve along which the surfaces intersect and if $\nu$ denotes the normal you need to show that along $\phi$: $ \nabla^{M_1}_{\phi'} \nu = \lambda \phi' \Rightarrow \nabla^{M_2}_{\phi'} \nu = \sigma \phi' $
where $\lambda$, $\sigma$ are scalar functions along $\phi$ and $\nabla^M$ denotes the derivative along the surface (i.e. the orthogonal projection of the derivative in the ambient space onto the tangent space of the surface). If you know how to compute this, the answer to i) should be quite obvious.
As for ii): convince yourself that the line of intersection is a curvature line in the plane. Since the surface touches the plane it is located in one of the halfspaces the plane defines. Write the surface locally as a graph of a nonnegative function $u$ over the plane, use the fact that $u$ has a minimum along the line of intersection. Check the implications of this observation for the second derivative and, in consequence, for the second fundamental form of the surface.