Abstract Algebra Computation 
Compute:
$3^{47}$ mod $23$
$3^{49}$ mod $7$
$2^{2^{17}}$ mod $23\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;$ hint: compute first $2^{17} + 1$ mod $19$

This is the first time I've had to compute elements, so I would appreciate any help.
For the first problem, I got $3^{47} \equiv 3^{22} \cdot 3^{22} \cdot 3^{3} \equiv 4$
For the second, I got $3^{49} \equiv (3^7)^7 \equiv 3^7 \equiv 3$
For the third, I'm not really sure where to begin. I don't understand how computing $2^17$ mod $18$ would help. Using Euler's theorem, though, I got that $2^{17}$ mod $18 \equiv 12$ since $\phi(18) = 6$.
EDIT: I'm sorry, I mistyped the third problem. I will ask it in another question.

Even for the second, $\displaystyle3^3=27\equiv-1\pmod7,$
$\displaystyle\implies3^{49}=(3^3)^{16}\cdot3\equiv(-1)^{16}\cdot3\pmod7 $
For the third,
$\displaystyle2^{\left(2^{17}\right)}\equiv2^{\left(2^{17}\pmod{\phi(23)}\right)}\pmod{23}$
Let us find $\displaystyle2^{17}\pmod{22}$ as $\phi(23)=22$
As $\displaystyle(2^{17},22)=2,$ we shall try finding  $\displaystyle2^{16}\pmod{11}$
As $\displaystyle2^5=32\equiv-1\pmod{11},2^{16}=2\cdot(2^5)^3\equiv2\cdot(-1)^3\pmod{11}\equiv-2\equiv9$
$\displaystyle\implies2^{17}\equiv2\cdot9\pmod{2\cdot11}\equiv18$
$\displaystyle\implies2^{\left(2^{17}\right)}\equiv2^{18}\pmod{23}$
Now, $\displaystyle2^6=64\equiv-5\pmod{23}$
$\displaystyle\implies2^{18}\equiv(-5)^3\equiv-125\equiv13$
