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 $e^x$.
Considering the equation$$A = M \frac{\left(1+\frac{R}{N}\right)^{NT}-1}{\frac RN}\left(1+\frac{R}{N}\right)\tag1$$ define
$$a=\frac AM \qquad, \qquad x=\frac RN\qquad,\qquad n=NT$$ to make the equation
$$a=\frac{(1+x)^n-1} x(1+x) \tag 2$$ What we know is that $x\ll 1$; so, let us develop the rhs using the binomial theorem
$$a=n+\sum_{k=1}^\infty \binom{n+1}{k+1}x^k$$ Transform it as simple Padé approximants which could be
$$a=n\frac{ (n+5) x+6}{6-2 (n-1) x}\implies x=\frac{6 (a-n)}{2 a (n-1)+n (n+5)}$$
Let us try with $a=500$ and $n=200$. The above formula would give
$$a=\frac{3}{400}=0.00750$$ while the exact solution would be $0.00809$; this is not so bad taking into account the huge value assigned to $a$.
Better, but at the price of a quadratic equation
$$a=n\frac{60 +6 (n+13) x+ (n^2+3 n+20) x^2 } {60-24 (n-2) x+3 (n-2) (n-1) x^2 }$$ For the worked example, selecting the "reasonable" root, this would give $x=0.00812$
For sure, we could continue improving but this would be at the price of cubic or quartic equations which can be solved with radicals. To give you an idea, using cubic equations, we should get $x=0.00809422$ while the exact solution is $0.00809450$.
Play with that and, please, tell me how it works for your cases.
Edit
The first way described above is "neutral" in the sense that $a$ is set equal to the ratio of polynomials leading to linear, quadratic, cubic or quartic equations in $x$.
There is another way. Rewrite $(2)$ as $(3)$
$$\frac 1a=\frac{x}{(x+1) \left((x+1)^n-1\right)}\tag 3$$ Expand the rhs as a Taylor series centered at $x=0$ and use series reversion to get
$$x=t+\frac {b_1} 6 t^2+\frac {b_2} {36} t^3+\frac {b_3} {1080} t^4+\frac {b_4} {6480} t^5+\frac {b_5} {90720} t^6+\frac {b_6} {2721600} t^7+O(t^{8})$$ where $t=\frac{2(a-n)}{(n+1)a}$. The coefficients are listed in the table below
$$\left(
\begin{array}{cc}
k & b_k \\
1 & n+5 \\
2 & 2 n^2+11 n+23 \\
3 & 22 n^3+153 n^2+402 n+503 \\
4 & 52 n^4+428 n^3+1437 n^2+2438 n+2125 \\
5 & 300 n^5+2836 n^4+11381 n^3+24879 n^2+30911 n+20413 \\
6 & 3824 n^6+40692 n^5+188712 n^4+496259 n^3+799917 n^2+780417 n+411779
\end{array}
\right)$$
Defining
$$x_{(p)}=t+\sum_{k=0}^p \frac{b_k}{c_k}\,t^{k+1}$$ for the worked example we should get the following values
$$\left(
\begin{array}{cc}
p & x_{(p)} \\
0 & 0.0059701493 \\
1 & 0.0071879409 \\
2 & 0.0076739522 \\
3 & 0.0078882746 \\
4 & 0.0079897311 \\
5 & 0.0080399577 \\
6 & 0.0080655906 \\
7 & 0.0080789649 \\
\cdots & \cdots \\
\infty &0.0080945103
\end{array}
\right)$$
Best Answer
You can make a quick determination using the following inequality which holds for all $n \in \mathbb{N}$ and $r > -n$,
$$e^r \left(1 - \frac{r^2}{n}\right) \leqslant \left( 1 + \frac{r}{n}\right)^n \leqslant e^r$$
If $r = 5\%$, then $ \left( 1 + \frac{r}{n}\right)^n \geqslant e^r \left(1 - \frac{r^2}{n}\right)> 0.9 e^r$ holds for all $n >0.05^2/0.1 = 0.025$. In other words, the discretely compounded amount exceeds $0.9e^r$ in the case where $r = 5\%$ for any frequency.
Only with very high interest rates will this condition not hold for all $n \geqslant 1$. For example, if $r = 55\%$, then
$$\left( 1 + \frac{0.55}{1}\right)^1= 1.55 < 0.9e^{0.55}\approx 1.55998 < 1.625625 = \left( 1 + \frac{0.55}{2}\right)^2$$
In this case, we have $e^r \left(1 - \frac{r^2}{n}\right)> 0.9 e^r$ for $n > 0.55^2/0.1 = 3.025$. Because the bound is not extremely tight it determines that $0.9e^r$ is first exceeded when $n \geqslant 4$ when in fact this is true for $n \geqslant 2$. Nevertheless, it provides a reasonable guide.
Proof of inequality
Since $e^x \geqslant 1 +x$ and $e^{-x} \geqslant 1 - x$ for $x \geqslant 0$ we have for $r \geqslant -n$,
$$\left(1+ \frac{r}{n}\right)^n \leqslant e^{r}\leqslant \left(1- \frac{r}{n}\right)^{-n},$$
and
$$0 \leqslant e^{r} - \left(1+ \frac{r}{n}\right)^n = e^{r}\left[1 - e^{-r}\left(1+ \frac{r}{n}\right)^{n}\right]\\ \leqslant e^{r}\left[1 - \left(1- \frac{r}{n}\right)^{n}\left(1+ \frac{r}{n}\right)^{n}\right]\\= e^{r}\left[1 - \left(1- \frac{r^2}{n^2}\right)^{n}\right]$$
By Bernoulli's inequality, we have
$$\left(1- \frac{r^2}{n^2}\right)^{n}\geqslant 1 - n \cdot \frac{r^2}{n^2} = 1 - \frac{r^2}{n},$$
and it follows that
$$0 \leqslant e^{r} - \left(1+ \frac{r}{n}\right)^n \leqslant e^r \frac{r^2}{n}$$