The 1s orbital(s) do(es) not have any nodes. the 1s orbital(s) has(have) a node at the nucleus. the 3d orbital(s) has(have) a cloverleaf shape, with four lobes of electron density around the nucleus and two perpendicular nodal planes. the f orbitals are even more complex. the number of nodes (and nodal planes) depends on the specific orbital, but there will be more than for s, p, or d orbitals.

I actually couldn't guess at first what the question is. But looking closely at the statements, I deduced that some are correct and some are not. So, I think this is a true or false problem. So,

The 1s orbital(s) do(es) not have any nodes. - This is false. Nodes are the planes that the orbitals do not fill. The formula for the number of nodes is:

N = n - l
where 
n is the energy level
l is 0 for s subshell, 1 for p subshell, 2 for d subshell, 3 for f subshell; l also signifies the number of angular nodes.
Thus,
N = 1 - 0 = 1 node

The 1s orbital(s) has(have) a node at the nucleus. Since this is the opposite of the first statement, this is true.

The 3d orbital(s) has(have) a cloverleaf shape, with four lobes of electron density around the nucleus and two perpendicular nodal planes. 
This is true. The shape of d subshell is cloverleaf, and all have four lobes. Since l=2, there are 2 perpendicular or angular nodes.

The f orbitals are even more complex. This is true. The f subshell is the last subshell. It has complex shapes and it rarely comes up in chemistry.

The number of nodes (and nodal planes) depends on the specific orbital, but there will be more than for s, p, or d orbitals. This is false. In fact, f orbitals have more nodes because l = 3. That means they always have 3 angular nodes, which is greater than the other subshells.

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