Computing $7$th root of $2$ modulo $33$ 
*

*$\varphi(33) = 20$

*$ed = 1 \pmod {20}$, and $d$ is $7$, so $e$ is $3$. $(3 \times 7 \mod 20 \equiv 1)$


So $x^e \pmod {33}$ is the seventh root. But how do you compute $x$? 

Later: My error was thinking x needed to be computed; it was in fact given as 2.

if ed = 1 mod Phi(N), then raising to the dth power is the inverse of raising to the eth power. 
$(x^e)^d$ = x mod N, and $(x^d)^e$ = x mod N.
$x^d$ is the e-th root of x modulo N.
We know d is 7 and x is 2.  The value of e may be computed: 
$ed$ = 1 mod $\varphi(N)$, so $e * 7$ = 1 mod 20 = 1
$e * 7$ mod 20 = 1 so $e$ is 3.
And indeed, $2^3$ mod 20 is 1.  So the answer is $2^3$ or 8.
 A: We have to solve the equation 
$$x^7\equiv 2\ \pmod{33}$$
It is clear that $x$ must be coprime to $33$. Therefore (and because of $\phi(33)=20$), we have $$x^{21}\equiv 8 \equiv x \pmod{33}$$
So, the solution is $8$.
A: Based on your computation, $x^3$ is a $7$th root of $x$ modulo $33$ for any $x$ relatively prime to $33$ since
$$(x^3)^7 = x^{21} = x^{\varphi(20)} \cdot x = x \pmod{33}$$
(this is Euler's theorem). So  in your case, $2^3 = 8$ is the desired $7$th root.
As a quick check, $$8^7 = 2097152 = 33 \times 63550 + 2$$ as desired.
A: More generally, if $\gcd(k,\phi(n))=1$, then $$x^k\equiv a\pmod{n}\iff x\equiv a^{k^{-1}\bmod \phi(n)}\pmod{n}$$
In your case, let $(k,a,n)=(7,2,33)$. Then $7^{-1}\equiv 3\pmod{\phi(33)}$, so raise both sides of $x^7\equiv 2\pmod{33}$ to the power of $3$ (as the other answers have noticed) to get $x\equiv 2^3\equiv 8\pmod{33}$.

Another method to solve $x^k\equiv a\pmod{n}$ without necessarily knowing $\gcd(k,\phi(n))=1$ can be using primitive roots.
You're solving $x^7\equiv 2\pmod {33}$, i.e. the system $\begin{cases}x^7\equiv 2\pmod{3} \\x^7\equiv 2\pmod{11}\end{cases}$
Clearly $x^7\equiv 2\pmod{3}\iff x\equiv 2\pmod{3}$ (there are only the cases $x\equiv \{0,1,2\}\pmod{3}$ to check).
To solve $x^7\equiv 2\pmod{11}$, instead of checking $x\equiv \{0,1,2,\ldots,10\}\pmod{11}$ you can note that $2$ is a primitive root mod $11$, because $\gcd(2,11)=1$ and $\phi(11)=2\cdot 5$ and $2^5\not\equiv 1\pmod{11}$ and $2^2\not\equiv 1\pmod{11}$.
Let $x\equiv 2^k\pmod{11}$. Then you're solving $2^{7k-1}\equiv 1\pmod{11}$, i.e. $7k\equiv 1\equiv 21\pmod{10}$, i.e. $k\equiv 3\pmod{10}$, so $x\equiv 2^k\equiv 8\pmod{11}$.
$$\begin{cases}x^7\equiv 2\pmod{3} \\x^7\equiv 2\pmod{11}\end{cases}\iff \begin{cases}x\equiv 2\pmod{3}\\x\equiv 8\pmod{11}\end{cases}\iff x\equiv 8\pmod{33},$$
where in the last step you can apply the Chinese Remainder Theorem.
