First understand the method of image charges. The idea behind the method is to bypass actually solving the differential equation with boundary conditions, and instead "cheat" by guessing the correct solution. To this end, we find a configuration of imaginary charges that together with the real ones will make the potential on all surfaces be what is given.
In your case, you have two surface, each with constant potential zero. For a plane, if you have a charge $q$ at $(x,y,z)$ and another one at $(x,-y,z)$, clearly the potential at $y=0$ will be zero. For the sphere, if a charge is at radius $R$ from the sphere with radius $a$, an image charge with charge $-qa/R$ should be placed at radius $a^2/R$, but the same idea remains.
Now, first add the image charge for the sphere. Now the potential on the sphere is zero, but on the plane, we still have a gradient. However, if you mirror both charges off the plane, you should see that the potential on both the sphere and the plane is zero.
Writing this all down, we have:
$$\begin{eqnarray}
\phi({\bf x}) &=& \frac{q}{\sqrt{(x-R\cos\alpha)^2+(y-R\sin\alpha)^2+z^2}} \\
&-& \frac{q}{\sqrt{(x-R\cos\alpha)^2+(y+R\sin\alpha)^2+z^2}} \\
&-&\frac{qa}{R\sqrt{(x-(a^2/R)\cos\alpha)^2+(y-(a^2/R)\sin\alpha)^2+z^2}} \\
&+&\frac{qa}{R\sqrt{(x-(a^2/R)\cos\alpha)^2+(y+(a^2/R)\sin\alpha)^2+z^2}}
\end{eqnarray}$$
Clearly, at $y=0$ we have $\phi=0$. Now, what about on the hemisphere? points on the hemisphere satisfy $z^2 = a^2 - x^2-y^2$, so that substituting leads us to the potential on the hemisphere:
$$\begin{eqnarray}
\phi({\bf x}_{sphere}) &=& \frac{q}{\sqrt{-2xR\cos\alpha -2yR\sin\alpha + R^2 +a^2}} \\
&-& \frac{q}{\sqrt{-2xR\cos\alpha +2yR\sin\alpha + R^2 +a^2}} \\
&-&\frac{qa}{R\sqrt{-2x(a^2/R)\cos\alpha -2y(a^2/R)\sin\alpha + (a^2/R)^2 + a^2}} \\
&+&\frac{qa}{R\sqrt{-2x(a^2/R)\cos\alpha +2y(a^2/R)\sin\alpha + (a^2/R)^2 + a^2}} \\
&=&0
\end{eqnarray}$$
By putting the $a/R$ into the square root.
To write down the multipole expansion, you just need to write down the Taylor expansion of the potential around $1/r$, with $r = \sqrt{x^2+y^2+z^2}$. This gives that any expression of the form:
$$\lim_{r\to\infty}\frac{1}{\sqrt{r^2+b}}= \frac{1}{r} - \frac{b}{2r^3} + \frac{3b^2}{8r^5} + \cdots$$
You can then use this expansion on the components of the potential to get your result.
Quite simply, the dipole moment of a charged system depends on the coordinate origin. There is nothing particularly surprising or unphysical about this, and there are plenty of other quantities (such as orbital angular momentum) with that property. The dipole moment of a charged system is not a quantity that is defined for the system itself; it is only defined for the system in relation to a given coordinate system.
Best Answer
The key point is to make a distinction between the notions of "dipole" and "dipole moment". The latter is not the same thing as the presence of a dipole. In fact, given what is said regarding the dependence of dipole moment on the origin, it turns out you can have a nonzero dipole moment when there is only one point charge present - just take the origin as anywhere but the exact location of that point charge.
A dipole is, by definition, two separate and equal but opposite point charges. So no, you cannot have a dipole with unequal charges, by definition.
But the dipole moment refers to a specific term in an asymptotic expansion of the electrical potential at large distance, i.e. in mathematician's terms, the "expansion at infinity". It is called as such because a dipole always has a nonzero dipole moment (which is, as said in the book, also independent of the choice of origin) and, moreover, a "pure" dipole moment - i.e. where this term only is taken in the expansion, the potential described is that of an "ideal" dipole, meaning one in which we take the mathematical limit as the charge separation goes to zero while the charge magnitude goes to infinity, with both of these balanced against each other "just so" to end up in a finite and nonzero electric field configuration. Thus the term measures "how much 'ideal dipoleness'" there is in the field configuration in question. The more it dominates other terms, the more the field looks like an ideal dipole, at least very far away from the center.
To see that there is no need at all to have a dipole to have a dipole moment, simply note that the dipole moment of $N$ point charges is just
$$\mathbf{p} := \sum_{i=1}^{N} q_i \mathbf{r}_i$$
where $\mathbf{r}_i$ are the charge positions. For a solitary point charge, this is just
$$\mathbf{p} = q\mathbf{r}$$
and thus one charge anywhere but the origin, i.e. $\mathbf{r} \ne \mathbf{0}$, has a nonzero dipole moment, yet is quite obviously about as far from being a "dipole", as just defined, as you can possibly get. The reason that a dipole - i.e. without "moment" after it - has a dipole moment independent of the origin is that if you take the above formula with two charges, i.e. $N = 2$, and equal and opposite, i.e. $q_1 = q_2 = q$, you get that the dipole moment becomes $q(\mathbf{r}_1 - \mathbf{r}_2)$, and while a position vector depends on the choice of origin, the difference of two such does not, because effectively, it is measuring the directed "distance" between two points, and intuitively, simply moving your reference point doesn't make things get closer or further apart from each other! If the two charges are slightly unbalanced, i.e. the scenario you are thinking of, you can think of the resulting dipole moment as this term for the mean charge, plus some unbalancing terms that depend on the position vectors individually, and so upon the origin. The dipole moment will never be outright zero in this case, and it will have a minimum, but it will also not be constant with respect to changes in the origin. Moreover, with regard to what we said about the asymptotic expansion, there will be other terms than just the dipole moment present in these cases - in fact, even for a "non-ideal" dipole - which account for the differences from the ideal dipole situation and, given the strong differences in field, these terms will be comparable if not dominant contributors in terms of their associated coefficients.