(1) For the first part of the problem, we will compare $\log_{2}3$ and $\log_{3}4$:

$\log_{2}3 - \log_{3}4 = \frac{\log 3}{\log 2} - \frac{\log 4}{\log 3} = \frac{\log^{2}3 - \log 2 \log 4}{\log 2 \log 3}$.

Applying the AM-GM inequality, we have:

$\log^{2}3 - \log 2 \log 4 > \log^{2}3 - (\frac{\log 2 + \log 4}{2})^{2} > \log^{2}3 - (\frac{1}{2}\log 9)^{2} = 0$.

So, $\log_{2}3 > \log_{3}4$.

Similarly, we can prove that $\log_{3}4 > \log_{4}5$.

The general conclusion is that $\log_{n}(n+1) > \log_{n+1}(n+2)$ where $n \in \mathbb{N^{+}}$.

To prove this, we have:

$\frac{\log_{n+1}(n+2)}{\log_{n}(n+1)} = \log_{n+1}(n+2)\log_{n+1}n 0$, we can conclude that $\log_{n}(n+1) > \log_{n+1}(n+2)$ for all $n \in \mathbb{N^{+}}$.

(2) For the second part of the problem, we first prove that $(a^{2}+b^{2})(x^{2}+y^{2}) \geq (ax+by)^{2}$:

Expanding the equation, we get:

$(a^{2}+b^{2})(x^{2}+y^{2}) - (ax+by)^{2} = a^{2}y^{2} - 2abxy + b^{2}x^{2} = (ay - bx)^{2} \geq 0$.

Since $(ay - bx)^{2} \geq 0$, we can conclude that $(a^{2}+b^{2})(x^{2}+y^{2}) \geq (ax+by)^{2}$.

Now, let's apply this conclusion to find the minimum value of $(\sin^{2}x+\cos^{2}x)(\frac{1}{\sin^{2}x}+\frac{4}{\cos^{2}x})$:

Applying the proved inequality, we have:

$(\sin^{2}x+\cos^{2}x)(\frac{1}{\sin^{2}x}+\frac{4}{\cos^{2}x}) \geq 9$.

\boxed{The minimum value of (\sin^{2}x+\cos^{2}x)(\frac{1}{\sin^{2}x}+\frac{4}{\cos^{2}x}) is 9.}