The solutions are the ones you listed.
The solutions all have shape $y=(x-c)^2$ or $y=0$. Thus if $b<0$, then none of the solutions curves pass through $(a,b)$. So for all pairs $(a,b)$ such that $b<0$, there cannot be a solution satisfying $y(a)=b$. We do not know (yet) whether these are all the pairs $(a,b)$ for which there is no solution, but soon we will.
For any $a$, if $b=0$ there are exactly two solutions satisfying $y(a)=b$, the singular solution and the solution $y=(x-a)^2$.
Finally, we look at pairs $(a,b)$ with $b$ positive. We look for values of $c$ such that $y(a)=b$.
The solution $y=(x-c)^2$ passes through $(a,b)$ if and only if $(a-c)^2=b$. This equation has exactly two solutions, $c=a\pm\sqrt{b}$.
Conclusion: (a) The pairs $(a,b)$ for which there is no solution satisfying $y(a)=b$ are all $(a,b)$ with $b<0$. (b) There are no pairs $(a,b)$ for which there are infinitely many solutions with initial condition $y(a)=b$. (c) For all remaining pairs $(a,b)$, that is, all pairs with $b \ge 0$, there are finitely many solutions, indeed exactly two solutions that satisfy $y(a)=b$.
The geometry: The conclusion can also be reached geometrically, by visualizing the family of parabolas. All of your parabolas are obtained by sliding the standard parabola $y=x^2$ along the $x$-axis. For any $(a,b)$ with $b \gt 0$, there are exactly two such parabolas that pass through (a,b): one whose "left" half goes through $(a,b)$, and one whose "right" half goes through $(a,b)$.
Note: One could interpret the word "finite" to include the possibility of $0$ solutions: $0$ is certainly finite! That is obviously not the intended interpretation here. But if we interpret "finite" as including $0$, the answer to (c) is all pairs $(a,b)$.
The indicial roots are $0$, $-1$, $-2$. There is a series solution
of the form
$\sum_{k=0}^\infty c_{2k} x^{2k}$ with $c_0 = 1$, $c_2 = -a/8 - 1/24$, and
$$ -(n+5) b c_n + ((a+1)n + 5a + 3) c_{n+2} +
+ (n+4)(n+5)(n+6) c_{n+4} = 0 $$
and a series solution of the form $\sum_{k=0}^\infty c_{2k-1} x^{2k-1}$ with $c_{-1}=1$,
$c_1 = -a/3$, and this same recurrence. A third fundamental solution involves $x^n$ for even $n \ge -2$ and $x^n \ln(x)$ for even $n \ge 0$.
Best Answer
Answer: For this you have to check the uniqueness of solution (click here for the theorem) of the given differential solution. If there is unique solution then, the given differential equation does not have any singular solution. If your given differential equation is of Clairaut's form , i.e., of the form $$y=px+f(p)\qquad \text{where $p=\frac{dy}{dx}$}$$ then you make a conclusion that this differential equation has singular solution (for example click here).
Answer: Yes, linear differential equations does not have any singular solutions. For explanation click here.
Answer: Yes, there may exists more than one singular solutions. For counter example click here.