Unless I did a mistake in the calculations, this should solve the last one.
$$24^{4493}=2^{3*4493}*3^{4493}$$
Now using
$$2^7 \geq 5^3 \,;\, 3^3 \geq 5^2 \,,$$
we get
$$2^{3*4493}*3^{4493} \geq (2^7)^{1925}(3^3)^{1497} \geq 5^{1925*3+1497*2}=5^{8769}$$
And here is the other
$$(\frac{79}{81})^{20}= (1-\frac{2}{81})^{20} \geq 1-\frac{40}{81} \,.$$
by Bernoulli
Thus
$$(\frac{79}{81})^{100}\geq \frac{1}{2^5} \geq \frac{1}{79} \,.$$
Thus
$$79^{101} \geq 81^{100} \,.$$
And hence
$$79^{1212} \geq 81^{1200} $$
The positivity of the second term is an immediate consequence of this....
P.S. Edit The last inequality also follows by this idea:
We show that
$$24^{4493} \geq 25^{4096}$$
$$(\frac{24}{25})^{12} \geq 1-\frac{12}{25} \geq \frac{1}{2}$$
$$(\frac{24}{25})^{12*342}\geq \frac{1}{2^{342}} \geq{1}{24^{389}} \,.$$