The state $\psi(x) = \delta(x)$ is a perfectly valid state for the harmonic oscillator to occupy. (With caveats, though: it is not normalizable, so it's not a physically-accessible state. Still, it's a perfectly reasonable thing for the mathematical formalism to handle.) As you note, it has a position uncertainty equal to zero, as well as a vanishing expectation value $⟨x^2⟩=0$.
However, it does not have a vanishing momentum uncertainty, and in fact if you expand it as a superposition of plane waves,
$$
\delta(x) = \frac{1}{2\pi\hbar} \int_{-\infty}^\infty e^{ipx/\hbar}\mathrm dp = \frac{1}{\sqrt{2\pi\hbar}} \int_{-\infty}^\infty A(p) e^{ipx/\hbar}\mathrm dp,
$$
you require an even weight $A(p) \equiv 1/\sqrt{2\pi\hbar}$ for all momenta, which means that the momentum-squared expectation value
$$
⟨p^2⟩ = \int_{-\infty}^\infty |A(p)|^2 p^2\mathrm dp = \frac{1}{2\pi\hbar}\int_{-\infty}^\infty p^2\mathrm dp = \infty
$$
diverges to infinity. (This result is required by the uncertainty principle, but the derivation here does not rely on it - it's an independent proof of that fact. Still, you can see the consistency in that $\Delta x=0$ and $\Delta p \geq \hbar/2\Delta x$ can only be satisfied by having $\Delta p = \infty$.)
This then implies that the expectation value of the hamiltonian is also infinity:
$$
⟨H⟩ = \frac{1}{2m}⟨p^2⟩ + \frac12 k ⟨x^2⟩ = \infty.
$$
As for this,
In the analysis of Harmonic Oscillator, it is claimed that $\langle\hat H\rangle$ cannot be zero, why is it so?
this is the zero-point energy of the oscillator, which has been explored multiple times on this site. If you want to ask why this is, you should ask separately, with a good showing of the previous questions here and what it is about them you do not understand.
