Solution:
a.
Let $x$ be the expected number of flips. There is a $\frac{1}{4}$ chance that she gets two heads right away. There is a $\frac{1}{2}$ chance the first flip is tails, in which case she is basically starting over after the first flip, so it will take an expected $x+1$ flips total. Finally, there is a $\frac{1}{4}$ chance the first flip is heads and the second tails, in which case she is basically starting over after 2 flips, so it will take an expected $x+2$ flips total. Thus we get
$$x = \frac{1}{4} \cdot 2 + \frac{1}{2}(x+1) + \frac{1}{4}(x+2) = \frac{3}{2} + \frac{3}{4} x.$$
Solving this gives $x = 6$.

b.
Let the expected number of flips be $y$. There is a $\frac{1}{2}$ chance you get tails on the first flip, in which case she is basically starting over, so it will take an expected $y+1$ flips total. If she gets heads on the first flip, then she is just waiting until she first gets tails. She has a $\frac{1}{2}$ chance of getting tails on each flip, so it will be an expected 2 flips before she first gets tails, for an expected 3 flips total. Thus we get
$$y = \frac{1}{2}(y+1) + \frac{1}{2} \cdot 3 = \frac{1}{2} y + 2.$$
Solving this gives $y = 4$.