Let's calculate:

- $$(5h \ 3 ' \ 34 '' )-(1h \ 35 ' \ 21 '' )$$
- $$(6h \ 7 ' \ 55 '' )\cdot 5$$

### Development:

1. Since the minutes to subtract $$(35)$$ are bigger than those that we have $$(3)$$ we must subtract one hour from the original time and add $$60$$ minutes to the minutes of this first time. This leads us to the operation to be calculated as:

$$(4h \ 63 ' \ 34 '' )-(1h \ 35 ' \ 21 '' )$$

Once we have the expressed the operation like this, we proceed to subtract each of the factors separately and we have:

$$3h \ 28 ' \ 13 ''$$

2. We multiply every factor by $$5$$ and we get: $$30h \ 35' \ 275''$$

Since we have more than $$60$$ seconds, we convert them into minutes. We divide $$275$$ by $$60$$. We take the integer part.

$$\dfrac{275}{60}=4,5833$$

We multiply the integer part $$4$$ by $$60$$ and we subtract it from $$275$$ original. We get:

$$4 \cdot 60 = 240 \Rightarrow 275-240=35$$

Therefore, we must add $$4$$ more minutes to the result and leave only $$35$$ seconds. This is: $$30h \ 39' \ 35'' $$

### Solution:

- $$3h \ 28' \ 13'' $$
- $$30h \ 39' \ 35'' $$