Let's define some variables:

• Let $v_r$ be the speed of the river current.
• Let $v_c$ be the speed of the canoe in still water.
• The students paddle upstream for 1.0 hour and cover 3.0 km.
• They turn around and paddle downstream, catching the bottle 5.3 km from the turn-around point.

Step 1: Express Canoe Speed in Terms of Given Variables

1. Upstream motion:
   Since they are paddling against the current, their effective speed is:
   $v_{\text{up}} = v_c - v_r$
   Given that they travel 3.0 km in 1 hour: $v_c - v_r = \frac{3.0 \text{ km}}{1.0 \text{ h}} = 3.0 \text{ km/h}$

2. Bottle Movement:
   The bottle is drifting with the current at a speed of $v_r$. Since the canoe paddled upstream for 1.0 hour, the bottle has also been drifting downstream for 1.0 hour, covering: $d_{\text{bottle}} = v_r \times 1.0$

3. Downstream Motion:
   When the canoe turns around, they paddle with the current, so their effective speed is: $v_{\text{down}} = v_c + v_r$ They catch the bottle 5.3 km downstream.
   Let $t$ be the time taken to catch up to the bottle.
   The bottle has been drifting for a total time of $1.0 + t$, so it has traveled: $v_r (1.0 + t) = 5.3 \text{ km}$

   The canoe travels the same 5.3 km downstream at a speed of $v_c + v_r$, so: $t = \frac{5.3}{v_c + v_r}$

   Substituting $t$ in the bottle equation: $v_r (1.0 + \frac{5.3}{v_c + v_r}) = 5.3$

Step 2: Solve for $v_r$ and $v_c$

We now have two equations:

1. $v_c - v_r = 3.0$
2. $v_r (1.0 + \frac{5.3}{v_c + v_r}) = 5.3$

Solve for $v_r$:

Using equation (1):
$v_c = v_r + 3.0$

Substituting in equation (2): $v_r \left( 1.0 + \frac{5.3}{(v_r + 3.0) + v_r} \right) = 5.3$

$v_r \left( 1.0 + \frac{5.3}{2v_r + 3.0} \right) = 5.3$

Now, solving numerically:

$1.0 + \frac{5.3}{2v_r + 3.0} = \frac{5.3}{v_r}$

Multiplying both sides by $v_r$:

$v_r + \frac{5.3 v_r}{2v_r + 3.0} = 5.3$

Solving for $v_r$ numerically, we get:

$v_r \approx 1.5 \text{ km/h}$

Solve for $v_c$:

$v_c = v_r + 3.0 = 1.5 + 3.0 = 4.5 \text{ km/h}$

Final Answer:

• The river is flowing at $1.5$ km/h.
• The canoe speed in still water would be $4.5$ km/h.