Recall that
[tex]{v_y}^2-{v_{0y}}^2=2a_y(y-y_0)[/tex]
At its maximum height [tex]y_{\mathrm{max}}[/tex], the toy will have 0 vertical velocity, so that
[tex]-\left(4.5\,\dfrac{\mathrm m}{\mathrm s}\right)^2=2\left(-9.8\,\dfrac{\mathrm m}{\mathrm s^2}\right)y_{\mathrm{max}}[/tex]
[tex]\implies y_{\mathrm{max}}=1.0\,\mathrm m[/tex]
For the toy to reach this maximum height, it takes time [tex]t[/tex] such that
[tex]\dfrac{0+v_0}2t=y_{\mathrm{max}}\implies t=0.46\,\mathrm s[/tex]
which means it takes twice this time, i.e. [tex]t=0.92\,\mathrm s[/tex], for the toy to reach its original position.
The velocity of the toy when it falls 1.0 m below its starting point is
[tex]{v_y}^2-\left(4.5\,\dfrac{\mathrm m}{\mathrm s}\right)^2=2\left(-9.8\,\dfrac{\mathrm m}{\mathrm s^2}\right)(0-1.0\,\mathrm m)[/tex]
[tex]\implies{v_y}^2=39.85\,\dfrac{\mathrm m^2}{\mathrm s^2}[/tex]
[tex]\implies v_y=-6.4\,\dfrac{\mathrm m}{\mathrm s}[/tex]
where we took the negative square root because we expect the toy to be moving in the downward direction.
