Are the following real numbers constructible? 1) $\sqrt[4]{5+\sqrt2}$
2)$\sqrt[6]{2}$
3) $3/(4+\sqrt13)$
4) $3+\sqrt[5]{8}$
From what I know, a number is constructible if it can be converted in a finite number of steps using only the operations addition, subtraction, multiplication, division, and square roots. 
This in mind, I would think #1 is constructible number because it is calculated using addition and square roots, with the 4th being a square root of a square root.
I would say #2 is not constructible as it is not possible to take a third root.
I would say #3 is constructible because it makes use of the addition, division, and square root operations.
Finally, #4 I would say is not constructible because it is not possible to generate the 5th root.
Can anyone weigh in on whether these conjectures are correct/incorrect, or maybe comment on the rationale and whether this method (if correct) could still be used on harder examples?
Thank you in advance!

While all these answers are correct, just having a higher root does not guarantee a failure of constructibility.  The number $\sqrt[3]{2+\sqrt5}\approx 1.618$ (big hint there) has been known to be constructible since Euclid, even if he would not have recognized the number in that form.
Suppose we are given a number $\sqrt[n]{a+\sqrt{b}}$, where $n$ is an odd natural number, $a$ is any integer and $b$ is any positive integer.  We can check such a number for cobstructibility with a two-step process.
First, if $\sqrt[n]{a+\sqrt{b}}$ is to be constructible then so is the conjugate $\sqrt[n]{a-\sqrt{b}}$.  Thus so is their product $\sqrt[n]{a^2-b}$ and thus, $a^2-b$ must be an $n$th power.  If this passes, define $\sqrt[n]{a^2-b}=R$ and move on to step 2.
In step 2, propose that
$(x+\sqrt{x^2-R})^n=a+\sqrt{b}.$
We also have the conjugate relation:
$(x-\sqrt{x^2-R})^n=a-\sqrt{b}.$
Clearly their product matches up with our defining equation $R^n=a^2-b$.  If you now expand the left sides by the binomial theorem and average the two results, the radicals cancel and you are left with
$x^n+(\frac{n(n-1)}{2})(x2-R)x^{n-2}+(\frac{n(n-1)(n-2)(n-3)}{24})(x^2-R)^2x^{n-4}+...+n(x^2-R)^{(n-1)/2}x=a$
If this has a rational root for $x$, then that rational root may be substituted for $x$ and we then have $x+\sqrt{x^2-R}$ as a constructibke $n$th root of $a+\sqrt{b}$.  Failure at either this step or step 1, above, means the numbervis not constructible.
Plug in $\sqrt[3]{2+\sqrt5}$ and see what happens.  First off $2^2-5=-1$ is a perfect cube, and accordingly we will set $R=-1$ for step 2.
For step 2, with a cube root the binomial expansion after canceling radicals will generate two terns, to wit:
$x^3+3(x^2+1)=2$
$4x^3+3x-2=0$
We seek a rational root, using the Rational Root Theorem to identify candidates.  In this case $x=1/2$ succeeds, thus our cube root is constructible and  we identify what we might have suspected from that decimal approximation I quoted:
$\sqrt[3]{2+\sqrt5}=(1/2)+\sqrt{(1/2)^2+1}=(1+\sqrt5)/2.$
