Answer
(a) $5.9 \times 10^5 \, \mathrm{N}$
(b) $1.8 \times 10^5 \, \mathrm{N} $
Work Step by Step
(a) The force exerted by the water on the bottom of the pool is simply equal to the weight of the water:
$$ F = mg = V \rho g = $$
$$ (4.0 \, \mathrm{m})(5.0 \, \mathrm{m})(3.0 \, \mathrm{m})(1.0 \times 10^3 \, \mathrm{kg/m^3})(9.8 \, \mathrm{m/s^2}) = 5.9 \times 10^5 \, \mathrm{N} $$
(b) I will use $x = 4.0 \, \mathrm{m}$ for pool width, $h$ for depth below the surface and $H = 3.0 \, \mathrm{m}$ for the total depth. I can now find the force $dF$ due to the pressure $dp$ on a thin, horizontal slice of the pool end with depth (or height) $dh$:
$$ dp = \frac{dF}{dA} = \rho gh \Rightarrow \frac{dF}{d(hl)} = \rho gh \Rightarrow dF = \rho glh \, dh $$
Integrating both sides (force goes from $0$ to $F$ when depth goes from $0$ to $H$) yields
$$ \int_0^F \, \mathrm{d}F = \rho gl \int_0^H h \, \mathrm{d}h \Rightarrow F = \rho gl \frac{H^2}{2} = $$
$$ (1.0 \times 10^3 \, \mathrm{kg/m^3})(9.80 \, \mathrm{m/s^2})(4.0 \, \mathrm{m}) \frac{(3.0 \, \mathrm{m})^2}{2} = 1.8 \times 10^5 \, \mathrm{N} $$
