Первая задача (проверил, вроде работает, константа N - количество чисел в последовательности):

const N = 10;
var number, count, i : integer;
begin
    count := 1;
    for i := 1 to N do
    begin
        write('Input number: ');
        readln(number);
        if (number mod 3 = 0) and (number 0) then
            count := count * number;
    end;
    if (count = 1) then
        count := 0;
    writeln('Result: ',count);
    readln;
end.

Вторая задача (проверил, работает, но мог недопонять условие):
const N = 10;
var number, positive, negative, i : integer;
begin
    negative := 1; positive := 0;
    for i := 1 to N do
    begin
        write('Input number: ');
        readln(number);
        if (number > 0) then
            positive := positive + number;
        if (number < 0) then
            negative := negative * number;
    end;
    writeln('Positive sum: ',positive);
    writeln('Negative mul: ',negative);
    if (positive > negative) then
        writeln('Sum of positive numbers bigger than mul of negative numbers.')
    else if (positive < negative) then
        writeln('Sum of positive numbers lesser than mul of negative numbers.')
    else
        writeln('Sum of positive numbers and mul of negative numbers are equal.');
    readln;
end.

Третья задача (проверил, работает):
const N = 10;
var number, lastneg, prevneg, i : integer;
begin
    lastneg := 100; prevneg := 100;
    for i := 1 to N do
    begin
        write('Input number: ');
        readln(number);
        if (number < 0) then
            if (lastneg > 0) then
                lastneg := number
            else
            begin
                prevneg := lastneg;
                lastneg := number;
            end;
    end;
    if (prevneg > 0) then
        writeln('This sequence of numbers contains less than two negative numbers.')
    else
        writeln('Penultimate negative number: ',prevneg);
    readln;
end.

Var
1 Задача
n, pr, i, a: integer;
begin
readln(n);
pr := 0;
for i := 1 to n do
begin
read(a);
if (a mod 3 = 0) then
if (pr > 0) then
pr := pr * a
else
pr := a;
end;
writeln(pr);
end.
2 Задача
var
n, i, pl, ot, a: integer;
begin
readln(n);
pl := 0;
ot := 0;
for i := 1  to n do
begin
read(a);
if (a >= 0) then
inc(pl, a)
else
begin
if (ot 0) then
ot := ot * a
else
ot := a;
end;
end;
if (pl > ot) then
writeln('Polozitelnii)
else
writeln('Otricat');
end.
var
n, i, a, posl, prd: integer;
begin
readln(n);
posl := 0;
prd := 0;
for i := 1 to n do
begin
read(a);
if (a < 0) then
begin
prd := posl;
posl := a;
end;
end;
if (posl = 0) then
writeln('Нет отрицательных')
else
writeln(prd);
end.