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Program ZAD00; {} uses crt;
var a,b:array[1..20] of real;
i:integer;
mina,minb,s:real;
Begin
{ заполнение массива А}
for i:=1 to 7 do begin write('введите элементы a[',i,']='); read(a[i]); end;
{ заполнение массива B}
for i:=1 to 9 do begin write('введите элементы b[',i,']='); read(b[i]); end;
{ min в массиве А}
mina:=a[1];
for i:=1 to 7 do begin
a[8]:= 32767 ;
if (a[i]>= a[i+1]) and (a[i+1]<=mina) then mina:=a[i+1]  end;
{ min в массиве B}
minb:=b[1];
for i:=1 to 9 do begin
b[10]:=32767;
if (b[i]>= b[i+1]) and (b[i+1]<=minb) then minb:=b[i+1]  end;
{ сложение minA и minB}
s:=mina+minb;
{ вывод результатов на экран
for i:=1 to 7 do begin write(a[i],' '); writeln; end;
for i:=1 to 9 do begin write(b[i],' '); writeln; end;
writeln('minA=',mina);
writeln('minB=',minb);
writeln('сумма их s=', s);
End.

Не трудно, но не полностью условие выполнено. Лови решение:

program arrs;
var
a : array[1..26] of integer;
i, min, max : integer;
m1, m2 : byte;

begin
randomize;
for i:=1 to 26 do begin
a[i] := random(29)-15;
write(a[i], ' ');
end;
min := 15;
max := -20;
for i:=1 to 26 do begin
if a[i] < min then begin
min:=a[i];
m1 := i;
end;
if a[i] > max then begin
max:=a[i];
m2 := i;
end;
end;
writeln;
writeln('Минимальный элемент в ячейке ', m1, ' равен ', min);
writeln('Максимальный элемент в ячейке ', m2, ' равен ', max);
end.

тестовый прогон:

0 2 -1 2 -9 -13 -9 -5 0 -14 11 6 8 -11 13 3 -15 -6 -15 -14 -3 13 9 7 3 -4
Минимальный элемент в ячейке 17 равен -15
Максимальный элемент в ячейке 15 равен 13