$\frac{(-1)^n}{2\cdot 4\cdot \cdot\cdot2n}=\frac{(-1)^n}{2^n\cdot n!}$ $$\frac{(-1)^n}{2\cdot 4\cdot \cdot\cdot2n}=\frac{(-1)^n}{2^n\cdot n!}$$
$$\frac{(-1)^n}{3\cdot 5\cdot \cdot \cdot(2n+1)}=\frac{{(-2)^n} \cdot n! }{(2n+1)!}$$
can anyone tell me if these are true or false?

Yes, both are true.
For the first, note that
$$ 2 \cdot 4 \dots 2n = (1 \cdot 2) \cdot (2 \cdot 2) \cdot (3 \cdot 2) \cdot (4 \cdot 2) \dots (n \cdot 2) = n! \cdot 2^n$$
For the second:
$$3 \cdot 5 \dots (2n + 1) = \frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \dots (2n + 1)}{2 \cdot 4 \dots 2n} = \frac{(2n+1)!}{n! \cdot 2^n}$$
and
$$2^n \cdot (-1)^n = (-2)^n$$
