Let the temperature of the surroundings be \( T_s \).

Given:
- First cooling: from \( 62^\circ \text{C} \) to \( 50^\circ \text{C} \) in 10 minutes.
- Second cooling: from \( 50^\circ \text{C} \) to \( 42^\circ \text{C} \) in the next 10 minutes.

Step 1: Use Newton's Law of Cooling
According to Newton's law:
\[
\frac{T_1 - T_2}{T_1 - T_s} = \frac{T_2 - T_3}{T_2 - T_s}
\]

Step 2: Substitute Values
Let \( T_1 = 62^\circ \text{C} \), \( T_2 = 50^\circ \text{C} \), and \( T_3 = 42^\circ \text{C} \).

Then:
\[
\frac{62 - 50}{62 - T_s} = \frac{50 - 42}{50 - T_s}
\]

Step 3: Solve for \( T_s \)
\[
\frac{12}{62 - T_s} = \frac{8}{50 - T_s}
\]

Cross-multiplying:
\[
12(50 - T_s) = 8(62 - T_s)
\]

Expanding and simplifying:
\[
600 - 12T_s = 496 - 8T_s
\]

\[
4T_s = 104 \Rightarrow T_s = 26^\circ \text{C}
\]

So, the temperature of the surroundings is \( 26^\circ \text{C} \).

Using Newton's law of cooling:

Given:
- Initial cooling: from \( T_1 = 61^\circ \text{C} \) to \( T_2 = 59^\circ \text{C} \) in \( \Delta t_1 = 10 \) minutes, with room temperature \( T_s = 30^\circ \text{C} \).
- New cooling required: from \( T_3 = 51^\circ \text{C} \) to \( T_4 = 49^\circ \text{C} \).

Using the proportionality from Newton's law:
\[
\frac{\Delta t_1}{\Delta t_2} = \frac{\text{Average Temp Difference in Initial Cooling}}{\text{Average Temp Difference in New Cooling}}
\]

Calculate average temperature differences:
1. For initial cooling: \( \frac{61 + 59}{2} - 30 = 60 - 30 = 30 \)
2. For new cooling: \( \frac{51 + 49}{2} - 30 = 50 - 30 = 20 \)

Then:
\[
\frac{10}{\Delta t_2} = \frac{30}{20}
\]

Solving for \( \Delta t_2 \):
\[
\Delta t_2 = 10 \times \frac{20}{30} = 15 \text{ minutes}
\]

Using Newton's law of cooling:

Given:
- Initial cooling: \( T_1 = 80^\circ \text{C} \) to \( T_2 = 79.9^\circ \text{C} \) in 5 min, surroundings \( T_s = 20^\circ \text{C} \).
- New cooling required: from \( 70^\circ \text{C} \) to \( 69.9^\circ \text{C} \).

For both cases, Newton's law states:
\[
\frac{\Delta T}{\Delta t} \propto (T - T_s)
\]

Since the temperature difference is small in both cases, we assume:
\[
\frac{\Delta t_1}{\Delta t_2} = \frac{T_1 - T_s}{T_3 - T_s}
\]

So:
\[
\frac{5}{\Delta t_2} = \frac{80 - 20}{70 - 20} \Rightarrow \Delta t_2 = 5 \times \frac{70 - 20}{80 - 20} = 5 \times \frac{50}{60} = 6 \text{ min}
\]

Using Newton's law of cooling:

Let \( T_s \) be the temperature of the surroundings, and \( T_1 = 70^\circ \text{C} \), \( T_2 = 60^\circ \text{C} \), \( T_3 = 54^\circ \text{C} \).

The average temperatures in each time interval are:
1. For first 5 minutes: \( \frac{70 + 60}{2} = 65^\circ \text{C} \)
2. For next 5 minutes: \( \frac{60 + 54}{2} = 57^\circ \text{C} \)

According to Newton's law:
\[
\frac{T_1 - T_2}{T_1 - T_s} = \frac{T_2 - T_3}{T_2 - T_s}
\]

Substitute values:
\[
\frac{70 - 60}{65 - T_s} = \frac{60 - 54}{57 - T_s}
\]

Simplifying:
\[
\frac{10}{65 - T_s} = \frac{6}{57 - T_s}
\]

Cross-multiplying and solving for \( T_s \) gives \( T_s = 45^\circ \text{C} \).