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 $e^x$.
To model your situation (in a ${continuous}$ fashion) you can use $y'=(a+bt)y$ which has a solution of $y(t)=y(0)e^{at+bt^2/2}$. To adjust $a$ and $b$ use
$\displaystyle \ln{y(t)\over y(0)}=at+bt^2/2$.
In your case $a=0.1$ and $t=3$ and the targeted growth is $y(3)/y(0)=2$. So $\ln(2)=0.1*3+b*3^2/2$
giving $b=(\ln(2)-0.1*3)*(2/3^2)=0.087$.
This again is for a continuous model. Companies act in step-wise/discrete fashion. In that case you have to use $y'=ay$ for one year, then $y'=(a+b)y$ for the next year, and then $y'=(a+2b)y$ in the next year, etc.
Best Answer
Do the start and end percentages mean the starting and ending values are like $(1+0\%)$ and $(1+2794.9\%)$ respectively? That will give $CAGR = 29.5\%$ by your formula.
The general idea seems to be that, imagine there's a base time before the start. In a timeline:
$$\text{Base}\to \text{Start}\ \underbrace{\to \cdots \to\cdots \to}_\text{many periods}\ \text {End}$$
The starting and ending values are both grown from the base value by the given percentages:
$$\begin{align*} \text{Starting value} &= \text{Base value} \cdot (1+\text{Starting percentage})\\ \text{Ending value} &= \text{Base value} \cdot (1+\text{Ending percentage})\\ \end{align*}$$
Then the goal is to find the average growth rate per period from the starting value to the ending value, or $CAGR$:
$$\begin{align*} \text{Ending value} &= \text{Starting value} \cdot (1+CAGR)^{\text{Number of periods}}\\ (1+\text{Ending percentage}) &= (1+\text{Starting percentage})\cdot (1+CAGR)^{\text{Number of periods}} \end{align*}$$