Yes, you can compute all the coefficients $e,f,g,E,F,G$ and get the gaussian and mean curvature and yes, it's tedious.
Here's another way:
From the first step we get : $Y_u\times Y_v=(1-2Ha+Ka^2)(X_u\times X_v)$, ie if $N$ and $\overline N$ are the normal vectors of $X$ and $Y$ respectively, then $\overline N\circ Y$ and $N\circ X$ coincide, since they're parallel. If these functions coincide then we have the following relations :
$$d\overline N(Y_u)=(\overline N\circ Y)_u=(N\circ X)_u=dN(X_u) \tag1$$
$$d\overline N(Y_v)=(\overline N\circ Y)_v=(N\circ X)_v=dN(X_v) \tag2$$
Let $\overline B$ be the matrix of $d\overline N$ with respect to $\{Y_u,Y_v\}$ and $B$ the matrix of $dN$ with respect to $\{X_u,X_v\}$.
Now, to compute $\overline K$ and $\overline H$ we need to find the expression of $\overline B$.
Put $$B=\begin{bmatrix}b_{11} & b_{12}\\ b_{21} & b_{22}\\ \end{bmatrix}$$
From the definition of $Y$ we have:
$$Y_u=X_u+a\cdot N_u=(a\cdot b_{11}+1)\cdot X_u+a\cdot b_{21}\cdot X_v$$
$$Y_v=X_v+a\cdot N_v=a\cdot b_{12}\cdot X_u+(a\cdot b_{22}+1)\cdot X_v$$
From these equations we can get the "change of basis" matrix : $Q=\begin{bmatrix}a\cdot b_{11}+1 & a\cdot b_{12}\\ a\cdot b_{21} & a\cdot b_{22}+1\\ \end{bmatrix}$ from $\{X_u,X_v\}$ to $\{Y_u,Y_v\}$. Then from the initial relations $(1)$ and $(2)$, we have the following equation:
$$B=Q\cdot \overline B$$
Since $Q$ is invertible: $$ \overline B=Q^{-1}\cdot B$$ From this point you can compute the entries of $\overline B$ and calculate $\overline H $ and $ \overline K$.
You can also notice that, since $Q^{-1}=(I+a\cdot B)^{-1}$, you have $\overline B=(I+a\cdot B)^{-1}\cdot B $. So, if $B$ has eigenvalues $-\lambda_1$ and $-\lambda_2$, then the eigenvalues of $\overline B$ are $\frac{-\lambda_1}{1-a\cdot \lambda_1}$ and $\frac{-\lambda_2}{1-a\cdot \lambda_2}$ and you can easily compute $\overline H$ and $\overline K$.
The key idea in this problem is to use an ODE comparison theorem, since you're given a curvature inequality. But first, we need to set up an appropriate ODE. Note that you're given a geodesic $\gamma$, and you want to show that there is a conjugate point in $\left(0, \frac{\pi}{\sqrt{\delta}} \right]$, which means you need to show that some non-trivial Jacobi field $J$ such that $J(0) = 0$ must vanish again before $\frac{\pi}{\sqrt{\delta}}$. Without loss of generality, assume that the Jacobi field $J$ is normal, i.e. perpendicular to $\gamma'$. Also assume that the geodesic $\gamma$ is parameterized by arc length. In this scenario, the Jacobi field $J(t)$ can be written in a nice manner (see Do Carmo exercise 1, ch. 5.5).
$$J(t) = \mu(t) e(t)$$
Here, $e(t)$ is the parallel transport of $J'(0)$ along $\gamma$, and $\mu(t)$ is just a smooth real valued function. The point of writing the
Jacobi field in this manner is to make the Jacobi equation particularly nice. It becomes the following.
$$\mu''(t) + K(t)\mu(t) = 0$$
Here, $K(t)$ is the sectional curvature of the plane spanned by $J(t)$ and
$\gamma'(t)$.
We know that $K(t)$ is bounded below by the constant function $\delta$. That means we can compare the above ODE to the following ODE.
$$
\mu''(t) + \delta\mu(t) = 0
$$
This ODE can be solved explicitly, and the solution to this vanishes
at $0$ and $\frac{\pi}{\sqrt{\delta}}$. A standard ODE comparison theorem then tells us that the roots of the solution to the previous ODE lie between $0$ and $\frac{\pi}{\sqrt{\delta}}$ (see Chapter 8 of Coddington's book on ODEs). This solves the problem.
EDIT: The ODE I had in the previous version was incorrect and didn't work. I replaced it with the correct ODE, and the arguments for the previously incorrect ODE work with the new correct ODE as well.
Best Answer
First of all, the answer to your question is easy: If you substitute $a=\dfrac1{2c}$ and $H=c$ into your equation, you get $$\frac K{1-2aH+a^2K} = \frac K{1-1+a^2K} = \frac1{a^2}=4c^2.$$
But be warned: The sign of $H$ depends on the parametrization of your surface (e.g., if you switch $u$ and $v$, the direction of the outward normal $N$ changes, and the sign of $H$ changes). So the "distance" of the parallel surface needs to be a signed distance. For example, with a sphere of radius $r$ with the usual outward-pointing normal, the principal curvatures are $-1/r$ and the mean curvature is $c=-1/r$. If you consider the parallel surface at distance $a$, this means you follow the normal a distance of $a$, which means your radius is now $r+a$, so the new principal curvatures are $-1/(r+a)$. But doCarmo wants you to use $a=\dfrac1{2c} = -\dfrac r2$, so we get $K=\left(-\dfrac1{r+a}\right)^2 = \left(-\dfrac1{r/2}\right)^2 = \dfrac4{r^2} = 4c^2$.