Here is the calculation for your first question. Start with a dollar. The nominal rate is $0.10$ per $9$ months, which I will take as meaning $\frac{3}{4}$ of a year. So the interest rate is $\frac{0.10}{3}$ per third of $9$ months, compounded every $3$ months.
So if we start with $1$ dollar, after $3$ months we have $\left(1+\frac{0.10}{3}\right)^1$, after $6$ months we have $\left(1+\frac{0.10}{3}\right)^2$, after $9$ months we have $\left(1+\frac{0.10}{3}\right)^4$. Finally, after one year we have $\left(1+\frac{0.10}{3}\right)^4$. Thus the effective annual interest rate is
$$\left(1+\frac{0.10}{3}\right)^4-1.$$
My calculator gives about $0.1401494$, a little bit over $14$%.
The calculation for your second question is mathematically very similar, but feels a little strange because of the unusual compounding.
The nominal interest rate is $0.10$ per $7$ months, compounded every $14$ months. So in $14$ months, $1$ dollar grows to $\left(1+\frac{0.10}{1/2}\right)$. (I am using this somewhat strange way of putting things, instead of writing $1+0.20$, so that you can fit it into the pattern of the formula.)
Now $1$ year is the fraction $\frac{12}{14}$ of the compounding period. So in one year, $1$ dollar grows to $\left(1+\frac{0.10}{1/2}\right)^{12/14}$, so the effective annual rate is
$$\left(1+\frac{0.10}{1/2}\right)^{12/14}-1.$$
The calculator gives an answer of about $0.1691484$.
The third question is the same, except easier. The effective annual rate is
$$\left(1+\frac{0.10}{1/2}\right)^{1/2}-1.$$
I hope these calculations are enough to tell you what's going on. Typically, that is in fact not how things are done. The usual way is to determine the "force of interest" and then use the exponential function $e^x$.
Best Answer
Note that in the United Kingdom where we have compound interest, consumer law requires APR and EAR interest rates to mean the same thing (apart perhaps from fees and other non-interest charges). I will use $r$ rather than your APR.
You are asking whether $(1+\frac rk)^k$ is an increasing function of positive $k$ (given fixed and presumably positive $r$)
and have found $f'(k)=(1+\frac rk)^k\left( \log_e(1+\frac rk) - \frac{r}{k+r}\right)$
which, as you note, is positive when $g(k)=\log_e(1+\frac rk) - \frac{r}{k+r} \ge 0$.
You could then say $\lim\limits_{k\to \infty} g(k)=0-0=0$ and $g'(k)=-\frac{r^2}{k(k+r)^2}<0$,
so $g(k)$ is decreasing towards $0$ as $k$ increases,
meaning $g(k)$ is positive for all finite positive $k$ and similarly $f'(k)$ is positive
and thus $f(k)$ is an increasing function of $k$.