$x^{311} \equiv 662 \mod{713}$ We have to find $x$ such that $x^{311} \equiv 662 \mod{713}$
The example given in my notes has a few typos and the professor is unavailable.

$$713=23 \cdot 31 $$
By Chinese Remainder Theorem, you need to solve
$$x^{311} \equiv 662 \pmod{23} \\
x^{311} \equiv 662 \pmod{31}$$
or equivalently
$$x^{311} \equiv 18 \pmod{23} \\
x^{311} \equiv 11 \pmod{31}$$
It follows that $x$ is invertible in both arithmetics, and then by Fermat Little Theorem we have
$$x^3\equiv 18 \pmod{23} \Rightarrow x^{-1} \equiv x^{21} \equiv 18^7 \pmod{23}\Rightarrow x \equiv (18^7)^{-1} \pmod{23}$$
and
$$x^{11} \equiv 11 \pmod{31} \Rightarrow x^{3} \equiv x^{33} \equiv 11^{3} \pmod{31}\,.$$
Now use the CRT to find $x \pmod{713}$.
Second solution
As 662 is invertible, it follows that $x$ is invertible and hence, by Euler Theorem we have
$$x^{660} \equiv 1 \pmod {713} \,.$$
now, find $a,b$ so that 
$$a\cdot 311 +b \cdot 660=1$$
Then
$$x= x^{1}=x^{a\cdot 311 +b \cdot 660}=(x^{311})^a\cdot(x^{660})^b=662^a \pmod{713} \,.$$
