A finite sequence $\{a_n\}$ has 50 terms and is such that $a_{n+1} = a_n + 0.15$ for $n = 1,2,3,\ldots,49$. Another infinite sequence $\{b_m\}$ is such that $b_1 = a_{50}$ and $\dfrac{b_m}{b_{m-1}} = 0.98$ for $m \ge 2$.

1. Identify the sequence as an arithmetic progression and express $a_{50}$ in terms of $a_1$.

Since $a_{n+1} - a_n = 0.15$ is a constant, $\{a_n\}$ is an arithmetic sequence with a common difference $d = 0.15$.
The 50th term is given by the formula $a_n = a_1 + (n-1)d$:
$a_{50} = a_1 + (50 - 1)(0.15)$
$a_{50} = a_1 + 49(0.15)$

2. Equate to the given relation and solve for $a_1$.

Substitute $a_{50}$ into the given equation $a_{50} = 99a_1$:
$a_1 + 49(0.15) = 99a_1$
$98a_1 = 49(0.15)$
Divide both sides by $49$:
$2a_1 = 0.15$
$a_1 = 0.075 \text{ (shown)}$

1. Apply the sum formula for an arithmetic progression.

The sequence is an arithmetic progression with $n = 50$. Using the sum formula $S_n = \frac{n}{2}(a_1 + a_n)$:
$\sum_{n=1}^{50} a_n = \frac{50}{2} (a_1 + a_{50})$

2. Substitute the relation $a_{50} = 99a_1$ and $a_1 = 0.075$ to evaluate the sum without a calculator.

Substitute $a_{50} = 99a_1$ (given in part (i)):
$\sum_{n=1}^{50} a_n = 25(a_1 + 99a_1)$
$= 25(100a_1)$
Substitute $a_1 = 0.075$:
$= 25(100 \times 0.075)$
$= 25 \times 7.5 = 187.5$

(Alternatively, using $S_n = \frac{n}{2}[2a_1 + (n-1)d]$:
$S_{50} = 25[2(0.075) + 49(0.15)] = 25[0.15 + 49(0.15)] = 25[50(0.15)] = 25(7.5) = 187.5$)

1. Calculate the values of $a_{25}$ and $b_1$.

$a_{25} = a_1 + 24d$
$a_{25} = 0.075 + 24(0.15) = 3.675$

$b_1 = a_{50} = 99a_1$
$b_1 = 99(0.075) = 7.425$

2. Formulate the inequality $b_k < a_{25}$ using the $n$th term formula for a geometric sequence.

$b_k < a_{25}$
$b_1(r)^{k-1} < a_{25}$
$7.425(0.98)^{k-1} < 3.675$
$(0.98)^{k-1} < \frac{3.675}{7.425}$
$(0.98)^{k-1} < \frac{49}{99}$

3. Solve for the smallest integer $k$ using logarithms.

Taking the natural logarithm on both sides:
$(k-1) \ln(0.98) < \ln\left(\frac{49}{99}\right)$

Since $\ln(0.98)$ is negative, dividing by it reverses the inequality sign:
$k - 1 > \frac{\ln(49/99)}{\ln(0.98)}$
$k - 1 > 34.811$
$k > 35.811$

The smallest integer value of $k$ is $36$.

1. Set up an inequality comparing the sum of the first $h$ terms ($S_h$) to the sum to infinity ($S_\infty$).

$S_h > 0.99 S_\infty$
$\frac{b_1(1 - 0.98^h)}{1 - 0.98} > 0.99 \left( \frac{b_1}{1 - 0.98} \right)$

2. Simplify the inequality by cancelling common factors.

Since the first term $b_1 = a_{50} = 0.075 + 49(0.15) = 7.425$ is positive, we can divide both sides by $\frac{b_1}{1 - 0.98}$ without reversing the inequality sign:
$1 - 0.98^h > 0.99$
$0.98^h < 0.01$

3. Solve the exponential inequality using logarithms.

Taking the natural logarithm on both sides:
$\ln(0.98^h) < \ln(0.01)$
$h \ln(0.98) < \ln(0.01)$
Since $\ln(0.98)$ is negative, dividing by it reverses the inequality sign:
$h > \frac{\ln(0.01)}{\ln(0.98)}$
$h > 227.95$

4. State the least integer value for $h$.

Since $h$ represents the number of terms, it must be an integer.
The least value of $h = 228$.

1. Determine the first term and common ratio of the new series

From earlier parts, $a_1 = 0.075$ and the common difference for $a_n$ is $0.15$.
$b_1 = a_{50} = a_1 + 49d = 0.075 + 49(0.15) = 7.425$

The series to evaluate is $\sum_{m=0}^{\infty} b_{1+3m} = b_1 + b_4 + b_7 + \dots$
This is a new geometric series with the first term $A = b_1 = 7.425$.
Since the consecutive ratio of $b_m$ is $r = -0.98$, the common ratio for this new series is taking jumps of 3 terms:
$R = r^3 = (-0.98)^3 = -0.941192$

2. Calculate the sum to infinity of the new series

$\begin{aligned} S_\infty &= \frac{A}{1 - R} \\ &= \frac{7.425}{1 - (-0.98)^3} \\ &= \frac{7.425}{1 + 0.941192} \\ &= \frac{7.425}{1.941192} \\ &\approx 3.82497 \end{aligned}$

$\therefore \sum_{m=0}^{\infty} b_{1+3m} = 3.82 \text{ (to 3 s.f.)}$