The proof of the asymptotic formula
$$
\Phi(n) = \frac{3}{\pi^2} n^2 + O(n\log n)
$$
can be modified to give a simple upper bound that has the right asymptotic character (but is not very accurate). We'll follow the proof in Chandrasekharan's Introduction to Analytic Number Theory.
By interpreting $\Phi(n)$ as the number of lattice points with relatively prime coordinates in or on the triangle $0 < y \leq x \leq n$, Chandrasekharan obtains the formula
$$
\Phi(n) = \frac{\Psi(n)+1}{2},
$$
where
$$
\Psi(n) = \sum_{1 \leq d \leq n} \mu(d) \left\lfloor \frac{n}{d} \right\rfloor^2.
$$
If we write
$$
\left\lfloor \frac{n}{d} \right\rfloor = \frac{n}{d} - \left\{\frac{n}{d}\right\},
$$
then
$$
\begin{align}
\Psi(n) &= \sum_{1 \leq d \leq n} \mu(d)\left(\frac{n}{d} - \left\{\frac{n}{d}\right\}\right)^2 \\
&= n^2 \sum_{1 \leq d \leq n} \frac{\mu(d)}{d^2} - 2n \sum_{1 \leq d \leq n} \frac{\mu(d)}{d}\left\{\frac{n}{d}\right\} + \sum_{1 \leq d \leq n} \mu(d) \left\{\frac{n}{d}\right\}^2.
\end{align}
$$
Now
$$
\begin{align}
-2n \sum_{1 \leq d \leq n} \frac{\mu(d)}{d}\left\{\frac{n}{d}\right\} &< 2n \sum_{1 \leq d \leq n} \frac{1}{d} \\
&< 2n \left(1+\int_1^n \frac{du}{u}\right) \\
&= 2n(1+\log n)
\end{align}
$$
and
$$
\sum_{1 \leq d \leq n} \mu(d) \left\{\frac{n}{d}\right\}^2 < \sum_{1 \leq d \leq n} 1 = n,
$$
giving us
$$
\Psi(n) < n^2 \sum_{1 \leq d \leq n} \frac{\mu(d)}{d^2} + 2n\log n + 3n.
$$
For the sum we get
$$
\begin{align}
\sum_{1 \leq d \leq n} \frac{\mu(d)}{d^2} &= \sum_{1 \leq d \leq
\infty} \frac{\mu(d)}{d^2} - \sum_{n+1 \leq d \leq \infty} \frac{\mu(d)}{d^2} \\
&= \frac{6}{\pi^2} - \sum_{n+1 \leq d \leq \infty} \frac{\mu(d)}{d^2} \\
&< \frac{6}{\pi^2} + \sum_{n+1 \leq d \leq \infty} \frac{1}{d^2} \\
&< \frac{6}{\pi^2} + \int_n^\infty \frac{du}{u^2} \\
&= \frac{6}{\pi^2} + \frac{1}{n},
\end{align}
$$
so that, in total,
$$
\Psi(n) < \frac{6}{\pi^2} n^2 + 2n\log n + 4n
$$
and hence
$$
\Phi(n) < \frac{3}{\pi^2} n^2 + n\log n + 2n + \frac{1}{2}.
$$
From Wikipedia: if $\displaystyle n=p_1^{k_1}\cdots p_r^{k_r}$, then
$\varphi(n)=\varphi(p_1^{k_1})\varphi(p_2^{k_2})\cdots\varphi(p_r^{k_r})=p_1^{k_1}\left(1- \frac{1}{p_1}\right)p_2^{k_2}\left(1-\frac{1}{p_2}\right)\cdots p_r^{k_r}\left(1-\frac{1}{p_r} \right)=$
$=n\cdot\left(1-\frac{1}{p_1}\right)\left(1-\frac{1}{p_2}\right)\cdots\left(1-\frac{1}{p_r}\right)$
So you'll have to find all different prime factors $p_1,\cdots,p_r$ of n.
Best Answer
$36=4\times9$
Since $4$ and $9$ are relatively prime, we have
$\phi(36)=\phi(4)\phi(9)$
Using the identity that for any prime $p$, $\phi(p^n)=p^n-p^{n-1}$
$\phi(4)\phi(9)=\phi(2^2)\phi(3^2)=(2^2-2).(3^2-3)=12$
As for the second part of the question, then, 13788 (mod 12) (mod 36) $\ne 138(mod 36)$ and hence your question is wrong.
There seems to be typo error and this part of the question can be modified to $13^{788}(mod\ 36) = 13^{788(mod \phi(36))}(mod\ 36) = 13^{788(mod 12)}(mod\ 36)=13^8(mod\ 36)=25$