Calculating the cut-off frequency
You calculated \$A_{ol}\$ well. You also know that the slope is 20 dB/dec, so with a \$G_{BW}\$ of 1 MHz, the open loop amplification will be:
   10 (20 dB) @ 100 kHz
  100 (40 dB) @  10 kHz
 1000 (60 dB) @   1 kHz
10000 (80 dB) @ 100  Hz

At 10 Hz it should be 100000 (100 dB), but Aol = 86 dB = 20000 (You wrote mV/V first time, instead of V/mV). That means the gain is double that at 100 Hz, which means the frequency is half of 100 Hz, or 50 Hz. Or, in simpler terms:
$$\dfrac{f_{max}}{A_{ol}}=\dfrac{1\;\mathrm{MHz}}{20000}=50\;\mathrm{Hz}\tag{1}$$
Which means your result is right (save roundings), just that there are fewer steps. Then you ask for:

the values of output voltage if the frequencies are \$0.25f_{3\,dB}\$ and \$5f_{3\,dB}\$

but you don't specify for what input. Given your calculations afterwards, I'll assume you didn't mean the value of the output voltage, but the gain. Then, for the first one it's \$A_{ol}\$, because it can't amplify more than 86 dB, and for the second one it's:
$$20\log_{10}{\Biggl(A_{ol}\dfrac{1}{5}\Biggr)}=20\log_{10}{\Biggl(\dfrac{20000}{5}\Biggr)}\approx 72\;\mathrm{dB}\quad(4000)\tag{2}$$
So, you calculated it correctly, just that you went a longer way after it.