We can solve this by assuming some form for the graph. The computations may be easiest assuming that the graph of percentile points is piecewise linear.
Suppose the graph is the collection of line segments through the points $(0,0.560)$, $(0.5, 1.583)$, $(a, 3.000)$, $(b, 4.000)$, $(1, 6.660)$. This will automatically have the right minimum, maximum and median.
It is also easy to calculate the contributions of these regions to standard deviation, using a fact from calculus that if $y$ is a line segment from $(p_x,p_y)$ to $(q_x,q_y)$, then the integral of $y^2$ in that region is $(q_x-p_x)(p_y^2+p_y q_y+q_y^2)/3$.
So the requirements on the mean and mean-square of the population give
\begin{align}
1.793 = (0.5-0)&(0.560 + 1.583)/2\\
+\ (a-0.5)&(1.583+3.000)/2\\
+\ (b-a)&(3.000 + 4.000)/2\\
+\ (1-b)&(4.000 + 6.660)/2\\
\\
1.793^2+0.948^2 = (0.5-0)&(0.560^2 + 0.560\ 1.583 + 1.583^2)/3\\
+\ (a-0.5)&(1.583^2 + 1.583 \ 3.000 + 3.000^2)/3\\
+\ (b-a)&(3.000^2 + 3.000 \ 4.000 + 4.000^2)/3\\
+\ (1-b)&(4.000^2 + 4.000 \ 6.660 + 6.660^2)/3
\end{align}
This has the solution $a=0.944, b=0.976$, which looks like this graph:
We could similarly choose other shapes for the graph, or $y$ values other than $3.000$ and $4.000$ for the kinks in it. We can then determine the values of all the percentiles by interpolation from the corner points on the graph.