Your row reductions seem fine, thus far. You can row reduce some more: try to get an upper triangular matrix; short of that, it will simplify the calculation of the determinant.
$(a)$ You can factor out $4$ from the third row. $\text{ det} \begin{pmatrix} 3 & 2 & -1& 4 \\ 2 & k & 6 & 5 \\ 0 & 4 & 0 & 4 \\ 0 & 0 & 4 & -5\\ \end{pmatrix} =33\iff 4 \text{ det} \begin{pmatrix} 3 & 2 & -1& 4 \\ 2 & k & 6 & 5 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 4 & -5\\ \end{pmatrix} =33$
$ \text{Add}\;-2(R_3) \text{ to}\;R_1 \implies 4 \text{ det} \begin{pmatrix} 3 & 0 & -1& 2 \\ 2 & k & 6 & 5 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 4 & -5\\ \end{pmatrix} = 33$ $ \text{Add }\;R_4 \text{ to}\; R_2\implies 4\text{ det} \begin{pmatrix} 3&0&-1&2\\ 2&k&10&0\\ 0&1&0&1\\ 0&0&4&-5\\ \end{pmatrix} = 33$
At this point, I'd suggest simply expanding along the column containing $k$; with the additional row reduction, that may simplify the process. Don't worry if you end up with with an equation in which $k$ evaluates to a fraction (decimal)!
$ \text {det} \begin{pmatrix} 3 & 0 & -1& 2 \\ 2 & k & 10 & 0 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 4 & -5\\ \end{pmatrix} = \frac{33}{4}\tag{1}$ $k\text{ det}\; \begin{pmatrix} 3 &-1&2\\ 0&0&1\\ 0&4&-5\\ \end{pmatrix} - \text{ det}\; \begin{pmatrix} 3&-1&2\\ 2&10&0\\ 0&4&-5\\ \end{pmatrix}\tag{2}$ $= k\text{ det}\; \begin{pmatrix} 3 &-1&2\\ 0&0&1\\ 0&4&-5\\ \end{pmatrix} - \left(3\text{ det}\; \begin{pmatrix} 10&0\\ 4&-5\\ \end{pmatrix} -2\text{ det}\; \begin{pmatrix} -1&2\\ 4&-5\\ \end{pmatrix}\right)= \frac{33}{4} \tag{3}$ $-12k - [3(-50) - 2(-3)] = \frac{33}{4}$ $-12k +144 = \frac{33}{4}$ $k=\frac{181}{16}$
Note:
In $(1)$, I simply divided both sides of the equation by $4$.
In $(2)$, I expanded along the second column, the column containing $k$. Note the sign of each resulting determinants.
In $(3)$, I expanded along the first column of the second matrix in $2$; again, we need to keep.
The rest is simplification.