no, they can't. Consider a point $P$ outside of the cylinder, then any line tangent to the cylinder will lie on the 2 planes tangent to the cones passing through the point ( This can be noted by considering a projection on t o a plane perpendicular to the cylinder, then the tangent line remain a tangent line, but to a circle, in particular the tangency points are all on 2 lines, as the have been projected on to 2 points ). Now two lines of the tetrahedron must be on the same plane, and also a third must be in the same plane completing a triangle, but the only intersection of the plane is a line, but a line can't intersect a triangle on all side

Answer: no WLOG let the cylinder be the points of distance $1$ from the $z$-axis. Notice no vertex of the tetrahedron can lie on the cylinder as it would cause the tetrahedron to be degenerate and no two points can share the same $z$ component as it would cause their edge to not touch the cylinder. Let the projections of the four vertices onto the $xy$-plane be $P_1$, $P_2$, $P_3$, and $P_4$. Notice that lines $P_1P_2$, $P_1P_3$, $P_1P_4$, $P_2P_3$, $P_2P_4$, and $P_3P_4$ all must be tangent to the unit circle. Notice that $P_1$ only has two tangents to the unit circle, so by PHP it must be that WLOG that $P_1$, $P_2$, and $P_3$ are collinear. Similarly, for $P_1P_4$, $P_2P_4$, and $P_3P_4$ to all be tangent to the unit circle it must be the $P_4$ also lies on this common line. This would cause the tetrahedron to be degenerate, a contradiction.