Probable intended solution:

(1)

p=1;

for i=1:2022

   p=mod(p*2022,10000); 

end

disp(p)

yields the last four digits as 1584.

TI-84 plus CE calculator:

1 STO>P

For(I,1,2022)

(P*2022) STO>P

remainder(P,10000)STO>P

End

leaves 1584 in P.

(2)

Find the common log of the answer:

>> 2022*log(2022)/log(10)

ans =

          6684.28948783757

          

Then find the antilog of the mantissa:

>> 10^0.28948783757

ans =

          1.94754650808029

          

First four digits are 1947. In fact the accuacy should be reliable up to four fewer (for the four we lose to the characteristic) than show: 1.9475465080, although rounding would bring the last digit to 1; I'd say it does actually truncate to 0.  

However MATLAB can give us all 6685 digits:

sym(2022)^2022

gives an answer that begins

ans =

19475465080987987461884359219297025987247775482742134438609736103478324990332...

and ends

 ...2576971189936148072764791763047709211277938901049215288399825028158870438563056451584,

 

 with thousands of intervening digits.  As the logarithm above shows, there are 6685 digits altogether.