Let me basically repeat what R. says in the comments.

For 1, $\mathbb{Z}[[T]]$ and $\mathbb{Z}_p$ are solid becuase $\mathbb{Z}$ is solid and solid abelian groups are closed under all limits and colimits. But $\mathbb{R}$ is not solid, see 2.

For 2, the first point is that $\mathbb{R}$ is pseudocoherent as a condensed abelian group, i.e. the following equivalent conditions are satisfied:

- $Ext^i(\mathbb{R},-)$ commutes with filtered colimits for all i;
- $\mathbb{R}$ admits a projective resolution where the terms are
*compact* projective condensed abelian groups.

(The equivalence of 1 and 2 is valid in any abelian category generated by compact projectives. It's a good exercise in homological algebra if you'd like to try.)

The pseudocoherence of $\mathbb{R}$ follows from the short exact sequence $\mathbb{Z}\rightarrow\mathbb{R}\rightarrow\mathbb{R}/\mathbb{Z}$. Indeed, $\mathbb{Z}$ is clearly pseudo-coherent (it is projective and compact) and the Breen-Deligne resolution implies that any compact abelian group is psuedo-coherent. Equivalent condition 1) shows that pseudocoherent modules have the 2 out of 3 property in short exact sequences, so this implies that $\mathbb{R}$ is pseudo-coherent.

The second point is that for a pseudocoherent condensed abelian group $M$ the derived solidification of $M$ identifies with $\underline{RHom}(\underline{RHom}(M,\mathbb{Z}),\mathbb{Z})$. Indeed, using condition 2 one reduces to checking that when $M=\mathbb{Z}[S]$ for $S$ extr. disconnected we have that $\underline{RHom}(\mathbb{Z}[S],\mathbb{Z})$ lives in degree $0$ and $\underline{RHom}(-,\mathbb{Z})$ on it is the derived solidification of $\mathbb{Z}[S]$. But these we verified in producing the solid theory (they follow from Specker's theorem and the definition of solidification).

Thus it suffices to see that $\underline{RHom}(\mathbb{R},\mathbb{Z})=0$. But this was verified in the proof of the existence of the solid theory. It follows from the Breen-Deligne resolution and the fact that $\mathbb{R}$ is contractible (so its cohomology with $\mathbb{Z}$-coefficients vanishes).

For 3, the answer is yes (Edit: no, I misread and thought the question was about filtered *colimits*. See comments) . For ordinary tensor product this is clear, and it follows for derived tensor product because filtered colimits are exact so they have vanishing higher derived functors.