I'm not going to compute everything you requested but I'll show you the way:
A second order system has a transfer function in the following form:
$$
T(s)=\frac{C(s)}{R(s)}=\frac{\omega_n^2}{s^2+(2\zeta \omega_n)s+\omega_n^2}
$$
where \$C(s)\$ and \$R(s)\$ are output and input, respectively.
For a unit step input i.e. \$R(s)=1/s\$, the output will be:
$$
C(s)=\frac{1}{s} \cdot \frac{\omega_n^2}{s^2+(2\zeta \omega_n)s+\omega_n^2}
$$
You'll need to take the inverse Laplace transform of the expression above, and you'll get:
$$
c(t)=1-\frac{e^{-\zeta\omega_n\ t}}{\sqrt{1-\zeta^2}} \ \sin\Bigg(\omega_n \sqrt{1-\zeta^2}\ t+\tan^{-1}\Big(\frac{\sqrt{1-\zeta^2}}{\zeta}\Big)\Bigg)
$$
For the sake of simplicity and readability, we can make the following replacements and re-write the expression:
$$
\tan^{-1}\Big(\frac{\sqrt{1-\zeta^2}}{\zeta}\Big)=\phi \\
\omega_n\sqrt{1-\zeta^2}=\omega_d \\
\Rightarrow c(t)=1-\frac{e^{-\zeta\omega_n\ t}}{\sqrt{1-\zeta^2}} \ \sin(\omega_d \ t+\phi)
$$
The rest is simple: Make calculations for the following conditions:
- \$\zeta<1\$ i.e. underdamped
- \$\zeta=1\$ i.e. critically damped
- \$\zeta>1\$ i.e. overdamped
\$\zeta=0\$ i.e. oscillation is out of scope here
- Rise time has different definitions: Generally, it's the time taken by the output to reach 1. But it can be defined as from 10% to 90% for overdamped, and 5% to 95% for critically damped. To calculate rise time, calculate the time-delta for the required conditions by solving for t for c(t)'s limit values. For example, for underdamped, solve for t from \$c(t)=1\$.
- Settling time is the time taken by the output to reach its steady value with some error. It's constant for all the types: \$T_s\approx 4/(\zeta\omega_n)\$ for 2% error, and \$T_s\approx 3/(\zeta\omega_n)\$ for 5% error.
- Peak time is defined for underdamped response: The time taken by the output to reach its maximum i.e. overshoot. You can solve for t from \$dc(t)/dt=0\$ because the tangent at the peak is in parallel with the horizontal axis, so its slope is zero.
EDIT: \$\zeta=1\$ is a special condition because the denominator of the second term of \$c(t)\$ goes to infinity. So we should calculate for \$\zeta\rightarrow 1\$ instead (Hello there infinitesimal calculus!):
$$
\lim_{\zeta\rightarrow 1} c(t) = \lim_{\zeta\rightarrow 1} \Bigg(1-\frac{e^{-\zeta\omega_n\ t}}{\sqrt{1-\zeta^2}} \ \sin(\omega_d \ t+\phi)\Bigg)=1-e^{-\omega_n \ t} (1+\omega_n \ t)
$$
