1. The lifetime of a particle in its own reference system frame is $6*10^{-11} s$. In the laboratory system the particle travels 25 m before decaying.To make this experimentally possible what is the largest value of particle speed $v$ in the laboratory frame?

In relativity the 4-dimensional space-time interval $\sigma^2=x^2+y^2+z^2-c^2t^2$ remains unchanged in the particle reference frame and in the laboratory frame.

If $\tau=6*10^{-11}s$ is the lifetime of the particle in its own frame and $x=25 m$ is the space in the laboratory frame then

$-c^2\tau^2=x^2-c^2t^2$

where $t$ is the time in the laboratory frame.

$t=\sqrt{(x^2/c^2)+\tau^2}=\sqrt{\frac{25^2}{9*10^{16}}+36*10^{-22}}=8.333*10^{-8} s$

But from the relativity time dilation relation we have:

$t=\gamma*\tau$ that is $\gamma =t/\tau=1.3888.89s$

$\frac{1}{\sqrt{1-(v/c)^2}}=1388.89$

$1-(v/c)^2=5.184*10^{-7}$

$c-v=2.592*10^{-7}c=77.86 m/s$

2. Please find the gravitational frequency shift being given the gravitational potential.

The differential equation for the frequency shift is

$d\omega/\omega=-d\Phi/c^2$

where $d\omega$ is the infinitesimal (angular) frequency shift and $\phi=GM/R$ is the gravitational potential. Therefore

$\int_{\omega_0}^{\omega} (d\omega/\omega)=-(1/c^2)\int_{\phi_0}^{phi}d\phi=(GM/c^2)\int_{R_0}^R (dR/R^2)$

The solution is

$ln(\omega/\omega_0) =(GM/c^2)*[(1/R)-(1/R_0)]$

Reference

Gravitational Redshift (Wiki)