Once you have a parameterization of the surface, you can then manipulate it to eliminate the parameters and get an implicit equation for the surface. This is always possible for a quadric. For instance, a possible parameterization of the first cylinder is $$x = s \\
y = \sqrt{\frac{71}6} \cos t - \frac13 \sqrt{\frac{71}2} \sin t - \frac73 \\
z = \sqrt{\frac{71}6} \cos t + \frac13 \sqrt{\frac{71}2} \sin t + \frac43.$$ The parameter $s$ appears only in the first equation, so it can be eliminated immediately: $x$ can take any value and will not appear in the final equation. Adding the other two equations gives $$y+z = 2\sqrt{\frac{71}6}\cos t-1$$ or $$\cos t = \sqrt{\frac3{142}}(y+z+1).$$ Similarly, subtracting the first equation from the second gives $$\sin t = \frac1{\sqrt{142}}(3z-3y-11).$$ Substituting these values into $\cos^2t+\sin^2t=1$ and simplifying results in $$2y^2-2yz+2z^2+12y-10z-3=0.$$ A similar procedure can be performed for the second cylinder, although $s$ won’t be quite so easy to eliminate since that cylinder doesn’t parallel the $x$-axis. You can eliminate $t$ in the same way as above, which will leave you with a pair of polynomials in $s$. If they’re both quadratic, you use resultants to eliminate $s$. You can find an example of the latter in this answer. Otherwise, eliminating $s$ is a straightforward matter of back-substitution.
Instead of going through this process, however, you can construct the implicit equations of these cylinders directly. Start just as you might for a parameterization of the surface by computing the circle’s center and radius. The circle is specified as the intersection of a sphere with a plane, so the circle’s center is the orthogonal projection of the sphere’s center onto the plane: $\mathbf c=\left(\frac53,-\frac73,\frac43\right)^T$. Its radius can be found via the Pythagorean theorem: $r^2 = 25 - \frac43 = \frac{71}3$. (The distance between the sphere’s center and the circle’s center is $2/\sqrt3$.) All of the cylinders with this directrix are then affine images of the cylinder $x^2+y^2=\frac{71}3$. To compute the equations of these cylinders it’s convenient to switch to homogeneous coordinates and use matrix operations. The general equation of a quadric surface can be written in matrix form as $\mathbf x^TC\mathbf x=0$, where $C$ is a symmetric matrix. If $M$ is the matrix of a point transformation $\mathbf x'=M\mathbf x$, then $$\mathbf x^TC\mathbf x = (M^{-1}\mathbf x')^TC(M^{-1}\mathbf x') = \mathbf x'^T(M^{-T}CM^{-1})\mathbf x',$$ i.e., the quadric transforms as $M^{-T}CM^{-1}$.
Our base cylinder for these problems has the matrix $C=\operatorname{diag}(1,1,0,-71/3)$. Let $\mathbf v$ be the direction vector of the generatrix. Choose mutually-orthogonal unit vectors $\mathbf u_1$ and $\mathbf u_2$ parallel to the plane of the directrix circle and construct the transformation matrix $$M = \begin{bmatrix}\mathbf u_1 & \mathbf u_2 & \mathbf v & \mathbf c \\ 0&0&0&1\end{bmatrix}.$$ This transformation rigidly maps the $x$-$y$ plane onto the circle’s plane, translating the origin to its center, and maps the $z$-axis to a line through the circle’s center parallel to the generatrix. Then compute $M^{-T}CM^{-1}$, from which you can extract the implicit equation of the surface and rearrange or simplify it as needed. In principle, you could map the unit circle to the directrix instead—start with the unit cylinder $\operatorname{diag}(1,1,0,-1)$ and use $r\mathbf u_1$ and $r\mathbf u_2$ to construct $M$, but I think the calculations will usually be simpler the other way.
For example, in part (a) we have $\mathbf v=(1,0,0)^T$ and can take $\mathbf u_1=\frac1{\sqrt2}(-1,1,0)^T$ and $\mathbf u_2=\frac1{\sqrt{6}}(1,1,2)^T$: $$M = \begin{bmatrix}-\frac1{\sqrt2}&\frac1{\sqrt6}&1&\frac53 \\ \frac1{\sqrt2}&\frac1{\sqrt6}&0&-\frac73 \\ 0&\frac2{\sqrt6}&0&\frac43 \\ 0&0&0&1 \end{bmatrix}$$ with inverse $$M^{-1}=\begin{bmatrix}0&\sqrt2&-\frac1{\sqrt2}&3\sqrt2 \\ 0&0&\frac{\sqrt6}2&-\frac4{\sqrt6} \\ 1&1&-1&2 \\ 0&0&0&1 \end{bmatrix}.$$ The resulting matrix of the cylinder is $$\begin{bmatrix} 0&0&0&0 \\ 0&2&-1&6 \\ 0&-1&2&-5 \\ 0&6&-5&-3 \end{bmatrix},$$ yielding the equation $$2y^2-2yz+2z^2+12y-10z-3=0$$ as before. For part (b), you just have to replace the vector $\mathbf v$ in $M$ with $(-1,1,0)^T\times(0,0,1)^T=(1,1,0)^T$.
If you’re doing this by hand, it might be easier to break $M$ down into the product of a linear transformation and a translation $$M = \left[\begin{array}{c|c} L & \mathbf t \\ \hline \mathbf 0^T & 1 \end{array}\right] = \left[\begin{array}{c|c} I_3 & \mathbf t \\ \hline \mathbf 0^T & 1 \end{array}\right] \left[\begin{array}{c|c} L & \mathbf 0 \\ \hline \mathbf 0^T & 1 \end{array}\right]$$ so that $$M^{-1} = \left[\begin{array}{c|c} L^{-1} & \mathbf 0 \\ \hline \mathbf 0^T & 1 \end{array}\right] \left[\begin{array}{c|c} I_3 & -\mathbf t \\ \hline \mathbf 0^T & 1 \end{array}\right] = \left[\begin{array}{c|c} L^{-1} & -L^{-1}\mathbf t \\ \hline \mathbf 0^T & 1 \end{array}\right].$$ That way you only have to invert a $3\times3$ matrix instead of $4\times4$.
