Let assume that you consider unital algebra only (one can still study non unital algebra by unitarizing them, but notion of spectrum is always a little annoying when one want to consider non unital algebra) and that a $H^*$ algebra is a real $C^*$-algebra with morphism of real $C^*$-algebra $\mathbb{H} \rightarrow A$

First two remarks:

The name $H^*$ is IMHO not a good idea at all, first the $C$ in $C^*$-algebra is not for "complex", and second $H^*$-algebra are actually already a thing, see for example here. I would rather call them $\mathbb{H}$-$C^*$-algebra or maybe quaternionic $C^*$-algebras (as we say real $C^*$-algebras and not $R^*$-algebras)

This definition might not be completely satisfying in the sense that it does not generalizes complexe $C^*$-algebras: a complexe $C^*$-algebra is the same as a real $C^*$-algebra together with an injection of $\mathbb{C}$ into its center. but when you replace $\mathbb{C}$ by $\mathbb{H}$ you need to remove the assumption on the center (because $\mathbb{H}$ is non-commutative) and it is not clear that it is a good idea to just remove it. For example, it might be a good idea to add all sort of commutativity conditions that holds in $\mathbb{H}$, like " if $x$ is self adjoint then $x$ commute to all elements of $\mathbb{H}$ which would simplify some answer below...

The only part I don't know how to answer is the non-emptyness of the spectrum... I'm still thinking about it, and it might be related to those possible additional condition I'm mentioning just above. I will edit if I find something.

So, let $A$ be a quaternionic $C^*$-algebra, and $a \in A$ I guess you want to define the spectrum as $\{ h \in \mathbb{H} | a - h \}$ is non-invertible.

Then the proof that is compact is exactly the same as in the usual case: you prove that it is bounded and closed using that if $X$ is invertible and $Y$ is such that $\Vert Y \Vert < 1/\Vert X^{-1} \Vert $ then $X+Y$ is invertible using a series argument (that work for arbitrary complete normed algebra)

As I said, 'non-empty' is trickier because the spectrum in a real $C^*$-algebra don't have to be inhabited and we don't have holomorphic calculus at disposition for quaternionic $C^*$-algebra (althoug it might be an idea to develop it a little, the resolvant is still locally a formal series so maybe one can use a kind of quaternionic Liouville's theorem).
But it seems difficult to produce a counter example: For example, as soon as you have an element which commute to some purely imagnary unit $u$, then $x,x^*,u$ generates a complexe $C^*$-algebra (with $u$ as $i$) hence $x$ has a non-empty spectrum inside of it, and it implies that $x$ has a non-empty sepctrum (as in real $C^*$-algebra it also true that $x$ is invertible is a subalgebra if and only if it is invertible in the larger algebra: it is proved by taking the complex form of the algebras).

For your question 2, the usual proof caries over easily and one can see that $X$ is the space of $\mathbb{H}$-algebra morphism from $H(X)$ to $\mathbb{H}$ hence if $H( X) \simeq H(Y)$ then $X \simeq Y$ and the isomorphism between $H(X)$ and $H(Y)$ is induced by the homeomorphism between $X$ and $Y$..

For your question 3), you always have a morphism from $A$ to $H(Spec A)$ where $Spec A$ is the set of character of $A$ (i.e. morphisms of $\mathbb{H}$-algebra from $A$ to $\mathbb{H}$). So "the algebras for which this morphism is an isomorphism" is an answer to your question. of course one can hope to obtain something better... If I'm correct the quaternionic Stone-Weierstrass theorem work, and the only unital $\mathbb{H}$-*-sub-algebra of $H(X)$ which separates points is $H(X)$ itself, hence the comparison map $A \rightarrow H(Spec A)$ is always surjective.

But one can give a more explicit condition: A quaternionic $C^*$-algebra is of the form $H(X)$ if and only if its self adjoint element ($x^*=x$) are central (or just commute between themselves and to elements of $\mathbb{H}$).

Indeed, it is then obvious that the set of self-adjoint element is a real $C^*$-algebra with trivial involution hence that it is of the form $C(X,\mathbb{R})$ for some $X$.

Moreover, using a kind of polarization formula, one can write any element in a unique way in the form $x+iy+jz+kt$ where $x,y,z,t$ are self-adjoint, and because self-adjoint are central, those multiply exactly as function on $X$ with values in $H$.

11more comments