Energy of a photoelectron depends on which factors?

If ultraviolet radiation of $6.2 eV$ falls on an aluminium surface,then the kinetic energy of the fastest emitted electron is (work-function $= 4.2 eV$)

Photons of energy $7 \,eV$ are incident on two metals $A$ and $B$ with work functions $6 \,eV$ and $3 \,eV$,respectively. The minimum de-Broglie wavelengths of the emitted photoelectrons with maximum kinetic energies are $\lambda_A$ and $\lambda_B$ respectively,where $\lambda_A / \lambda_B$ is nearly

Statement $1$: When ultraviolet light is incident on a photocell, its stopping potential is $V_0$ and the maximum kinetic energy of photoelectrons is $K_{max}$. When $X$-rays are used instead of ultraviolet light, both $V_0$ and $K_{max}$ increase.
Statement $2$: Photoelectrons are emitted with a range of speeds from $0$ to a maximum value because the incident light contains a range of frequencies.

In a photoemissive cell with exciting wavelength $\lambda$,the fastest electron has speed $v$. If the exciting wavelength is changed to $\frac{3\lambda}{4}$,the speed of the fastest emitted electron will be

When a photon of energy $8 \ eV$ is incident on a metal surface with a threshold frequency of $1.6 \times 10^{15} \ Hz$,the maximum kinetic energy of the emitted photoelectrons is .......... $eV$. (Given: $h = 6.6 \times 10^{-34} \ Js, 1 \ eV = 1.6 \times 10^{-19} \ J$)