Step 1

$ L =200(10-t)^2$

Differentiating w.r.t $t$ we get

$ \large\frac{dl}{dt}=200 \times 2(10-t)(-1)$

$ = -400(10-t)$

when $t=5$

$ \large\frac{dL}{dt}=-400(5)$

$=-2000$

Hence the water is decreasing at the rate of $2000\: l/sec$

Step 2

When $t=5$

$ L=200(10-5)^2$

$=200 \times 25$

$ = 5000$

Hence when $ t=5$, the capacity of water is $5000$ litres.

Hence the water flowing out during the first 5 seconds

$ 5000-2000=3000 \: l$

Hence the capacity of water flowing out during the first 5 seconds is $ 3000 \: l$