Let's tackle each equation step by step. We'll first identify the center, foci, vertices, and lengths of the major and minor axes. For part (d), we will rewrite the given equation into a standard ellipse form by completing the square.

Part (a)

$\frac{(x - 1)^2}{100} + \frac{(y - 7)^2}{25} = 1$

This is the standard form of an ellipse equation $\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$, where $(h, k)$ is the center, $a^2$ is the square of the semi-major axis, and $b^2$ is the square of the semi-minor axis.

1. Center: $(h, k) = (1, 7)$
2. Vertices: Since $a^2 = 100$, $a = 10$. The vertices are 10 units from the center along the $x$-axis (since the larger number is under the $x$-term). So, the vertices are $(1 \pm 10, 7)$ or $(-9, 7)$ and $(11, 7)$.
3. Foci: To find the foci, calculate $c$ using the relation $c^2 = a^2 - b^2$. Here, $b^2 = 25$, so $c^2 = 100 - 25 = 75$, meaning $c = \sqrt{75} = 5\sqrt{3}$. The foci are located $c$ units from the center along the major axis (the $x$-axis), so the foci are $(1 \pm 5\sqrt{3}, 7)$.
4. Length of major axis: $2a = 20$.
5. Length of minor axis: $2b = 10$.

Part (b)

$\frac{(x - 2)^2}{36} + \frac{(y - 1)^2}{81} = 1$

Here, the larger number is under the $y$-term, so the major axis is along the $y$-axis.

1. Center: $(h, k) = (2, 1)$
2. Vertices: Since $a^2 = 81$, $a = 9$. The vertices are 9 units from the center along the $y$-axis, so the vertices are $(2, 1 \pm 9)$ or $(2, -8)$ and ((2