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 $e^x$.
The weighted averages a is not the correct approach. Here it is necessary to calculate the $\underline{\text{compounded}}$ annual return. After one period the you earn interest for the initial investment ($C_0$). Therefore after this period the initial investment increases to $C_1=(1+i_1)\cdot C_0$, where $i_k$ is the interest rate of the k-th period. And after n periods the growth of the investment $C_n=C_0\cdot \prod\limits_{k=1}^n (1+i_k)$. And the growth factor after $n$ periods is
$$g(n)=\frac{C_0\cdot \prod\limits_{k=1}^n (1+i_k)}{C_0}=\prod\limits_{k=1}^n (1+i_k)$$
To calculate the average growth rate we have to take the n-th root of $g(n)$ and then subtract 1. In other words, you obtain the compounded annual return. If you have a calculator it is not difficult to get the result. You just have to input $\left((1+i_1)\cdot (1+i_2)\cdot (1+i_3)\right)^{\frac13}$
Best Answer
The population of A after $t $ years is $100 (1+.1)^t$.
B is $100,000 (1+.01)^t $
C is $75,000 (1+.02)^t$
You can find when any two of them are equal by setting 2 equal to each other and solving for $t $. But there's no reason to believe there will be a time all three are equal. There might be. But there's no reason to assume there is.
So $A =B $ when $100*1.1^t= 100000*1.01^t $ or when $A/B=1$ so $\frac {100*1.1^t}{100000*1.01^t}=1$
So $\frac {100}{100000}(\frac {1.1}{1.01}^t)=1$
$10^{-3}(\frac {1.1}{1.01})^t=1$
$ (\frac {1.1}{1.01})^t=10^3$
$t=\log_{\frac {1.1}{1.01}}10^3$
$=\frac {\log_{10}10^3}{\log_{10}\frac {1.1}{1.01}} =\frac 3 {\log_{10}\frac {1.1}{1.01}}$