The maximum kinetic energy of emitted electrons in a photoelectric effect does not depend upon

In a photoelectric experiment,the stopping potential was measured to be $V_1$ and $V_2$ volts with incident light of wavelength $\lambda$ and $\frac{\lambda}{2}$ respectively. The value of $V_2$ is: [where $\phi=$ work function,$e=$ electronic charge]

Photoelectric emission is observed from a metallic surface for frequencies $v_1$ and $v_2$ of the incident light $(v_1 > v_2)$. If the maximum values of kinetic energy of the photoelectrons emitted in the two cases are in the ratio $1: n$,then the threshold frequency of the metallic surface is

Statement $-1$: When ultraviolet light is incident on a photocell, its stopping potential is $V_0$ and the maximum kinetic energy of the photoelectrons is $K_{max}$. When the ultraviolet light is replaced by $X$-rays, both $V_0$ and $K_{max}$ increase.
Statement $-2$: Photoelectrons are emitted with speeds ranging from zero to a maximum value because of the range of frequencies present in the incident light.

When radiation of frequency $8 \times 10^{15} \ Hz$ is incident on a metal surface with a work function of $6.125 \ eV$,what is the kinetic energy of the emitted photoelectrons in $eV$?

The photoelectric cut-off voltage in a certain experiment is $1.5 \; V$. What is the maximum kinetic energy of photoelectrons emitted?