Finding generating set for $\operatorname{Gal}_\mathbb{Q}f$ a) $f = x^3 + x^2 +1$ 
$f$ has no ration roots so its irreducible over $\mathbb{Q}$ and has discriminant =-31 so $\operatorname{Gal}_\mathbb{Q}f$ is isomorphic ot the full symmetric group $S_3$. Now $f$ has one real root, $a$, so in $\mathbb{Q}(a)$ f can be factored into irreducible polynomials as $f=(x-a)(x^2+(a+1)x+(a+a^2))$. Then take $g = (x^2+(a+1)x+(a+a^2))$ which is irreducible over $\mathbb{Q}(a)$ and you can find roots $= .5(-a-1 \pm \sqrt{-3a^2 -a +1})$.  Now in the example in my notes they get these are complex conjugates (I kind of see how that works because here a would be positive and bigger than one), but they also find a way to write $b^2$ in terms of $a$, which I can do by squaring both sides but this gets messy and I still have $a$ under the radical so I'm not sure this is how I want to proceed.  Is it?  Anyway, I think that from this I will get that $\operatorname{Gal}(\mathbb{Q}(a,b)/\mathbb{Q})$ is isomorphic to $\operatorname{Gal}_\mathbb{Q}(f)$ and the size of the Galois group = $[\mathbb{Q}(a,b):\mathbb{Q}]=6$ so then I know that the identity is an automorphism and there has to be at least one more that generates the Galois group. 
b) $f=x^4 + x^2 + 6$
This one I don't really know how to factor since it has only complex roots so its going to factor by letting $y=x^2 \rightarrow f=y^2 + y + 6 \rightarrow y=-1 \pm \sqrt{1-24} = -1 \pm i\sqrt{23} \rightarrow x = \pm \sqrt{-1 \pm i\sqrt{23}}$.  Obviously factoring $f$ becomes a nightmare.  Is there an easier way to solve this?  Or is what I'm doing not so bad and I'm just missing some trick?
 A: This started out as a comment, and grew monstrously.
I want to prove that your quartic polynomial is irreducible over $\Bbb Z$ (or over $\Bbb Q$, which is the same thing here). You have made a little error on identifying the roots: setting $y=x^2$, you correctly got $y^2+y+6=0$, but the roots of this are $\frac{-1\pm\sqrt{-23}}2$. You want to show that adjoining a root $x$ satisfying $f(x)=x^4+x^2+6=0$ induces a quartic extension of $\Bbb Q$, for this guarantees that $f(X)$ is irreducible.
Now we have $[\Bbb Q(x):\Bbb Q]=[\Bbb Q(x):\Bbb Q(y)][\Bbb Q(y):\Bbb Q]$, where the second factor on the right is definitely $2$. I claim that the other factor is also $2$, because if it were $1$, then $x=\sqrt y$ would be in $\Bbb Q(y)=\Bbb Q(\sqrt{-23}\,)$. But it’s easy to see that $y$ is not a square in the field $\Bbb Q(\sqrt{-23}\,)$, because its Norm down to $\Bbb Q$ is $6$, and if $\sqrt y\in\Bbb Q(y)$, then you’d have an element of this field whose Norm was $\sqrt6$ (the Norm is multiplicative). That’s impossible, so the upper level of the tower is also a quadratic extension.
Too much work for a relatively simple fact: very likely, others will give much shorter arguments. But the important upshot of irreducibility is that there is no $\Bbb Q$-factorization of $f(X)=X^4+X^2+6$: what you see is the only possible factorization. So you should not have been looking for any simpler factorization of this than the product of the four linears $(X-\rho)$, as $\rho$ runs through the four roots you (almost) wrote down.

EDIT — Addition:
In comments, you have wondered about the associated Galois group in part (b). Let’s be specific and let $\alpha$ be a root of $X^2+X+6$, which we may take to be $\frac{-1+\sqrt{-23}}2$, and $\bar\alpha$ its conjugate $\frac{-1-\sqrt{-23}}2$. Contiuing, let $\beta$ be a particular square root of $\alpha$, so that $\beta$ is a root of your quartic $f(X)=X^4+X^2+6$. Then the roots of $f$ are $\beta$, $-\beta$, and two others, call them $\gamma$ and $-\gamma$, with $\gamma^2=\bar\alpha$.
Let’s call $k_0=\Bbb Q(\alpha)$, $k=\Bbb Q(\beta)$, and $K=\Bbb Q(\beta,\gamma)$. Since $K$ has been gotten by adjoining all roots of $F$, it’s the splitting field, thus Galois over $\Bbb Q$. 
 What we know is that $\beta^2+\gamma^2=-1$, so that $\text{Irr}(\gamma,k[X])=X^2-(-1-\beta^2)=X^2+1+\beta^2$. (Note that $\gamma=\sqrt{\bar\alpha}\notin k$ for if not, $\alpha$ and $\bar\alpha$ would have to be related by a square factor in $k_0=\Bbb Q(\alpha)$ and they are not.)
Now I want to show that the Galois group $\text{Gal}^K_{\Bbb Q}$ is the dihedral group of order eight, $D_8$. I’ll do this by finding two automorphisms of $K$, $\sigma$ and $\tau$, which will satisfy $\tau^2=\text{id}$, $\sigma^4=\text{id}$, and $\tau\sigma\tau^{-1}=\sigma^{-1}$.
First, set $\sigma(\beta)=\gamma$, always possible to send one root of $f$ to any other. But this certainly means $\sigma(-\beta)=-\gamma$, so where can $\sigma$ send poor $\gamma$? To either one of $\pm\beta$, and I choose $\sigma(\gamma)=-\beta$. Then we have
$$\sigma: \beta\mapsto\gamma\mapsto-\beta\mapsto-\gamma\mapsto\beta\,,$$
which shows that $\sigma$ is of order $4$. For $\tau$ take the nontrivial $k$-automorphism of $K$, $\beta\mapsto\beta$, $\gamma\mapsto-\gamma$. Then $\tau^2=\text{id}$ and $\tau\sigma\tau$ maps $\beta\mapsto\beta\mapsto\gamma\mapsto-\gamma$, and this is $\sigma^{-1}$.
