The work done of a force when the force is $F_{0}$ acts over a distance $d_{0}$ is

$$
begin{align*}
W &= Fd \
W &= F_{0}d_{0}
end{align*}
$$

When the force is doubled $2F_{0}$ and the distance is halved $d_{0}/2$, the work becomes

$$
begin{align*}
W &= Fd \
&= left( 2F_{0} right) left( frac{d_{0}}{2} right) \
W &= F_{0}d_{0}
end{align*}
$$

The work $textbf{remains the same}$.

In this problem, a Darwin’s finch can exert a force of $F = 205~mathrm{N}$ over a distance of $d = 0.40 times 10^{-2}~mathrm{m}$. We calculate the work done by the finch.

From the definition of work, we have

$$
begin{align*}
W &= Fd \
&= left( 205~mathrm{N} right) left( 0.40 times 10^{-2}~mathrm{m} right) \
W &= boxed{ 0.82~mathrm{J} }
end{align*}
$$

$$
W = 0.82~mathrm{J}
$$

In this problem, a person lifts a pumpkin of mass $m = 3.2~mathrm{kg}$ to a height of $d = 0.80~mathrm{m}$ to check it out. We calculate the work done on the pumpkin to lift it from the ground. We use $g = 9.81~mathrm{m/s^{2}}$.

The force exerted by the person must be equal to the weight of the pumpkin, so $F = mg$. By the definition of work, we have

$$
begin{align*}
W &= Fd \
&= mgd \
&= left( 3.2~mathrm{kg} right) left( 9.81~mathrm{m/s^{2}} right) left( 0.80~mathrm{m} right) \
&= 25.11360~mathrm{J} \
W &= boxed{ 25~mathrm{J} }
end{align*}
$$

$$
W = 25~mathrm{J}
$$