From $textbf{Hook’s law}$, the force exerted by an ideal spring of constant $k$ is proportional to the distance of its stretch or compression $x$:

$$
begin{gather*}
F = kxtag{1}
end{gather*}
$$

The force exerted on the spring to stretch it is $F = 1.20$ N and the distance of stretch is $x = 6.40$ cm.

Simply, we plug our values for $F$ and $x$ into equation (1) and evaluate for $k$, so we get the spring constant for this spring:

$$
begin{align*}
(1.20text{ N}) &= (0.064text{ m}) k\\
k & = dfrac{1.20text{ N}}{0.064text{ m}} = 18.75text{ N/m}
end{align*}
$$

$$
therefore quad large boxed{k = 18.75text{ N/m}}
$$

From $textbf{Hook’s law}$, the force exerted by an ideal spring of constant $k$ is proportional to the distance of its stretch or compression $x$:

$$
begin{gather*}
F = kxtag{1}
end{gather*}
$$

The force exerted on the spring to stretch it is $F = 2.30$ N and the spring constant of the spring is $k = 19.0$ N/m.

Simply, we plug our values for $F$ and $k$ into equation (1) and evaluate for $x$, so we get the stretch of the spring:

$$
begin{align*}
(2.30text{ N}) &= (19.0text{ N/m}) x\\
x & = dfrac{2.30text{ N}}{19.0text{ N/m}} = 0.121text{ m}
end{align*}
$$

$$
therefore quad large boxed{x = 12.1text{ cm}}
$$