A) Find the focus, directrix, vertex and axis of symmetry for the parabola:
\[
8(y+2)=(x+1)^{2}
\]

Focus
Directrix
Vertex

Step 1 ：The given equation is in the form of \(4p(y-k) = (x-h)^2\), which is the standard form of a parabola that opens upwards or downwards.

Step 2 ：Here, \(h = -1\), \(k = -2\) and \(4p = 8\), so \(p = 2\).

Step 3 ：The vertex of the parabola is at the point \((h, k)\), the focus is at the point \((h, k+p)\), and the directrix is the line \(y = k - p\). The axis of symmetry is the vertical line \(x = h\).

Step 4 ：Substituting the values of \(h\), \(k\), and \(p\) into the formulas, we get the vertex at the point \((-1, -2)\), the focus at the point \((-1, 0)\), the directrix at the line \(y = -4\), and the axis of symmetry at the vertical line \(x = -1\).

Step 5 ：\(\boxed{\text{Final Answer: The vertex of the parabola is at the point } (-1, -2), \text{ the focus is at the point } (-1, 0), \text{ the directrix is the line } y = -4, \text{ and the axis of symmetry is the vertical line } x = -1}\)