In a multi-electron atom, the energy of an orbital depends mainly on: $n \text{ and } l$ because electron-electron repulsion and shielding remove the degeneracy that exists in hydrogen-like atoms.

In the absence of magnetic and electric fields, the energy does not depend on the magnetic quantum number $m$.

Therefore, orbitals having the same values of $n$ and $l$ but different $m$ will have the same energy.

Now examine the given orbitals:

• (A) $n=1,\ l=0,\ m=0$
• (B) $n=2,\ l=0,\ m=0$
• (C) $n=2,\ l=1,\ m=1$
• (D) $n=3,\ l=2,\ m=1$
• (E) $n=3,\ l=2,\ m=0$

Compare their $n$ and $l$ values:

• (D): $n=3,\ l=2$
• (E): $n=3,\ l=2$

These differ only in $m$, so they have the same energy.

Check other pairs:

• (B) and (C): same $n=2$, but $l=0$ and $l=1$ different $\Rightarrow$ different energy
• (C) and (D): different $n$ and $l$ $\Rightarrow$ different energy
• (A) and (B): different $n$ $\Rightarrow$ different energy

Hence the correct option is: $\boxed{\text{A: (D) and (E)}}$