Both facts have short proofs. We first prove (1) and then get a stronger statement for (2) - so that the result is $C^{\infty}$ smooth.

**Proof of (1)**. By assumptions, for each point $x$ of the set there is a half-space $H_x$ containing  $x$ in its boundary and containing all other points in its interior. So there is also a radius $R$ ball $B_x^R$ contained in $H_x$, tangent to $\partial H_x$ at $x$ and containing all the other points of the set (just take $R$ large enough). Now, the intersection of all balls $B_x^R$ is the desired strictly convex set.

**Proof of (2).** Let's show how to construct such a strictly convex function in the $C^{\infty}$ class. So suppose that our collection of points is $(x_i,y_i)$ and the planes are defined by $y=L_i(x)$ (so that $y_i=L_i(x)$). It is clear than the for $\varepsilon>0$ small enough the function 
$$F(x)=\max_i(L_i(x)+\varepsilon|x-x_i|^2)$$
solves the problem. However $F_{\varepsilon}$ is not smooth. To find a smooth function let's choose a symmetric density $\rho_{\varepsilon}(x)\ge 0$ satisfying 

1) $\rho_{\varepsilon }(x)=\rho_{\varepsilon }(-x)$, 2) $\int \rho_{\varepsilon }=1$, and 3) $\rho_{\varepsilon }(x)=0$ for $|x|>\varepsilon$.

The following claim is easy.

*Claim*. The convolution   $\widetilde F_{\varepsilon}=F_{\varepsilon}*\rho_{\varepsilon }$ has the property that $\widetilde F_{\varepsilon}-F_{\varepsilon}=const(\varepsilon)>0$ at points that are on distance more than $\varepsilon$ from the locus of non-smoothness of $F_{\varepsilon}$.

So it is easy to see that for $\varepsilon$ small enough the function $\widetilde F_{\varepsilon}-const(\varepsilon)$ will do the job.