Решите неравенства пожалуйста

Решите задачу:

$1) \ 5^{-x} \ \textgreater \ 625 \\ \\ 5^{-x} \ \textgreater \ 5^4 \\ \\ -x \ \textgreater \ 4 \\ \\ x \ \textless \ -4$

$2) \ \bigg (\dfrac{1}{5} \bigg )^{x^2 + x - 10} = 625 \\ \\ \bigg (\dfrac{1}{5} \bigg )^{x^2 + x - 10} = \bigg ( \dfrac{1}{5} \bigg )^{-4} \\ \\ x^2 + x - 10 = -4 \\ \\ x^2 + x - 6 = 0 \\ \\ x_1 + x_2 = -1\\ x_1 \cdot x_2 = -6 \\ \\ x_1 = -3 \\ x_2 = 2$

$3) \ 7^{x^2 - x + 3} \leq \bigg ( \dfrac{1}{7} \bigg ) ^{5x} \\ \\ 7^{x^2 - x + 3} \leq 7^{-5x} \\ \\ x^2 - x + 3 \leq -5x \\ \\ x^2 + 4x + 3 \leq 0 \\ \\ x^2 + 4x + 4 - 1 \leq 0 \\ \\ (x + 2)^2 - 1^2 \leq 0 \\ \\ (x + 2 - 1)(x + 2 + 1) \leq 0 \\ \\ (x + 1)(x + 3) \leq 0 \\ \\ x \in [-3; -1]$

$4) \ 9^x \cdot 27^y = 27 \\ 2^x/4^y = 32 \\ \\ 3^{2x} \cdot 3^{3y} = 3^{3} \\ 2^{x}/2^{2y} = 2^5 \\ \\ 3^{2x + 3y} = 3^{3} \\ 2^{x - 2y} = 2^{5} \\ \\ 2x + 3y = 3 \\ x - 2y = 5 \\ \\ 2x + 3y = 3 \\ 2x - 4y = 10 \\ \\ 7y = -7 \\ \\ y = -1 \\ 2x - 3 = 3 \\ \\ y = -1 \\ 2x = 6 \\ \\ y = -1 \\ x = 3 \\ \\ OTBET: \ (3; -1).$