A nominal rate of interest is always specified with two quantities: the annualized rate (typically expressed as a percentage), and the compounding frequency per year. So for example, if I say that a loan is repaid at a nominal rate of $i^{(6)} = 9\%$ compounded every two months (i.e., 6 times a year), then what this means is that every two months, interest of $i^{(6)}/6 = 1.5\%$ is accrued, and this is an effective two-month rate of interest.
The equivalent effective annual interest rate is therefore $$i = \left(1 + \frac{i^{(6)}}{6} \right)^6 - 1 = (1.015)^6 - 1 = 9.344\%.$$
At this same effective annual rate, what is the nominal rate compounded monthly; i.e., what is $i^{(12)}$? It is the solution to $$\left(1 + \frac{i^{12}}{12}\right)^{12} = \left(1 + \frac{i^{6}}{6}\right)^6,$$ or $i^{(12)} = 8.9665\%$. What is the nominal rate compounded daily (365 days/year)? $$i^{(365)} = 8.93426\%.$$ What is the nominal rate compounded continuously? it is $$\delta = i^{(\infty)} = 6 \log 1.015 \approx 0.0893317.$$ This is what we call the force of interest.
To answer your first question, 100 payable quarterly for 10 years at the end of each year, at a rate of $8\%$ convertible quarterly tells you that the $8\%$ figure is a nominal rate of interest because of the "convertible quarterly" phrase; that the effective quarterly rate is $i^{(4)}/4 = 2\%$; and the effective annual rate is therefore $i = (1.02)^4 - 1 = 8.243\%$. But we don't need the effective annual rate for our calculations because the payments are also made quarterly. If the payments were made for a period that doesn't correspond to the nominal conversion frequency, then we need to convert it to the equivalent effective rate for the period of payments.
Your question, "why even bother saying 'convertible quarterly,'" reveals exactly why the rate of $8\%$ is a nominal rate. If it were effective, as you observed, then there's no point in mentioning the conversion frequency.
$$\require{enclose}$$
Your confusion is precisely why I do not recommend memorizing all of the formulas for annuities. All you really need is to understand and memorize the basic formulas for the present and accumulated values of level annuities-immediate, -due; the increasing (and optionally decreasing) annuities; and the level annuities paid $m$ times per year. Geometrically increasing/decreasing annuities are simply level annuities with a modified interest rate. Increasing/decreasing annuities paid on a non-annual basis are equivalent to ones that are paid annually, again by converting the interest rate the effective $m^{\rm th}$ly rate. Over-reliance on formulas is the downfall of many students preparing for the actuarial exams.
Let us recall the following:
$$\begin{align}
i^{(m)} &= m((1+i)^{1/m} - 1) \tag{1} \\
v &= (1+i)^{-1} \tag{2} \\
a_{\enclose{actuarial}{n}i} &= \frac{1 - v^n}{i} \tag{3} \\
\ddot a_{\enclose{actuarial}{n}i} &= (1+i)a_{\enclose{actuarial}{n}i} \tag{4} \\
a_{\enclose{actuarial}{n}i}^{(m)} &= \frac{1}{m} a_{\enclose{actuarial}{mn}j} = \frac{1 - v^n}{i^{(m)}}, \quad j = i^{(m)}/m \tag{5} \\
(Ia)_{\enclose{actuarial}{n}i} &= \frac{\ddot a_{\enclose{actuarial}{n}i} - nv^n}{i}. \tag{6}
\end{align}$$
To understand what the scheulde of payments for the symbols $(Ia)_{\enclose{actuarial}{n}}^{(m)}$ and $(I^{(m)}a)^{(m)}_{\enclose{actuarial}{n}}$ look like, we have the table
$$\begin{array}{c|c|c|c|c|c}
\text{time} & a_{\enclose{actuarial}{n}} & a_{\enclose{actuarial}{n}}^{(m)} & (Ia)_{\enclose{actuarial}{n}} & (Ia)_{\enclose{actuarial}{n}}^{(m)} & (I^{(m)}a)^{(m)}_{\enclose{actuarial}{n}} \\
\hline
0 & 0 & 0 & 0 & 0 & 0\\
1/m & 0 & 1/m & 0 & 1/m & 1/m^2 \\
2/m & 0 & 1/m & 0 & 1/m & 2/m^2 \\
\ldots & \ldots & \ldots & \ldots & \ldots & \ldots \\
1 & 1 & 1/m & 1 & 1/m & 1/m = m/m^2\\
1 + 1/m & 0 & 1/m & 0 & 2/m & 1/m + 1/m^2 = (m+1)/m^2 \\
1 + 2/m & 0 & 1/m & 0 & 2/m & 1/m + 2/m^2 = (m+2)/m^2 \\
\ldots & \ldots & \ldots & \ldots & \ldots & \ldots \\
2 & 1 & 1/m & 2 & 2/m & 2/m = 2m/m^2 \\
\ldots & \ldots & \ldots & \ldots & \ldots & \ldots \\
(n-1) + 1/m & 0 & 1/m & 0 & n/m & (n-1)/m + 1/m^2 = (mn-m+1)/m^2 \\
(n-1) + 2/m & 0 & 1/m & 0 & n/m & (n-1)/m + 2/m^2 = (mn-m+2)/m^2 \\
\ldots & \ldots & \ldots & \ldots & \ldots & \ldots \\
n & 1 & 1/m & n & n/m & n/m = mn/m^2
\end{array}$$
Then it is clear that $(I^{(m)}a)^{(m)}_{\enclose{actuarial}{n}}$ is simply an increasing annuity of $1/m^2$ over a term of length $mn$ with periodic rate $j$; i.e.,
$$\begin{align}(I^{(m)}a)^{(m)}_{\enclose{actuarial}{n}}
&= \frac{1}{m^2} (Ia)_{\enclose{actuarial}{mn}j} \\
&= \frac{\ddot a_{\enclose{actuarial}{mn}j} - mn v^n}{m^2 j} \\
&= \frac{\frac{1}{m} \ddot a_{\enclose{actuarial}{mn}j} - n v^n}{i^{(m)}} \\
&= \frac{\ddot a_{\enclose{actuarial}{n}}^{(m)} - nv^n}{i^{(m)}}. \tag{7}
\end{align}$$
This proves the formula in your question.
Now, the next part pertains to how formula $(7)$ is applied to the specific problem. In this case, we have $n = 5$ years, $m = 12$ payments per year (i.e., monthly). We also have that the first payment is $2$ at the end of the first month and increases by $2$ each month. Since the first payment of the symbol $(I^{(m)}a)^{(m)}_{\enclose{actuarial}{n}}$ is $1/m^2$, it follows that we must multiply this by $2(12^2) = 288$, as you correctly deduced. Finally, we are given that $i^{(4)} = 0.09$, which means $$i = \left(1 + \frac{i^{(4)}}{4}\right)^4 - 1 \approx 0.0930833$$ and $v = (1+i)^{-1} \approx 0.914843$. We also have $$i^{(12)} \approx 0.0893333.$$ So far, we are in agreement. However,
$$\ddot a_{\enclose{actuarial}{5}}^{(12)} = (1+i)^{1/12} \frac{1 - v^5}{i^{(12)}} \approx (1.0930833)^{1/12} \frac{1 - (1.0930833)^{-5}}{0.0893333} \approx 4.05064401427.$$ Your calculation is incorrect because you must divide your result by $12$, as the monthly payments are each $1/12^{\rm th}$ of the equivalent annual payment over $60$ periods.
Finally, we have
$$288(I^{(12)}a)^{(12)}_{\enclose{actuarial}{5}} \approx 288 \frac{4.05064401427 - 5(0.914843)^5}{0.0893333} \approx 2729.21$$
as claimed.
But as I warned at the very beginning, this is all entirely unnecessary. There is too much needless memorization and too much opportunity to make an error. The simplest approach is to scale the time period accordingly: the effective periodic interest rate is $j = i^{(12)}/12 \approx 0.00744444$, hence the present value is simply
$$2(Ia)_{\enclose{actuarial}{60}j} = 2\frac{\ddot a_{\enclose{actuarial}{60}j} - 60v_j^{60}}{j},$$ where all we need to do is compute $v_j = (1+j)^{-1} \approx 0.992611$ and $$\ddot a_{\enclose{actuarial}{60}j} = 48.6077.$$ We get the same result with less complexity and fewer formulas to memorize.
Best Answer
I don´t know what went wrong. Let a,b the lower bound and upper bound, respectively. And $m=\frac{a+b}2$. Then I have the following decision rule:
It results in the following table:
It is necessary that the root (j) is between $x_1$ and $x_2$. If it is not then the bisection method went wrong. In your case $j=0.025244$ is outside of the interval $(0.0406, 0.0559)$