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
$24^{4493} \geq 25^{4096}$
$(\frac{24}{25})^{12} \geq 1-\frac{12}{25} \geq \frac{1}{2}$
$(\frac{24}{25})^{12*342}\geq \frac{1}{2^{342}} \geq{1}{24^{389}} \,.$