Recheck your calculation. I get $$\Pr[V \mid S] = \frac{(0.04)(0.95)}{(0.04)(0.95) + (0.96)(0.01)} = \frac{95}{119} \approx 0.798319.$$
As for your other question, you want $$\Pr[V^c \mid S^c].$$ This is solved in an analogous fashion to what you already computed:
$$\Pr[V^c \mid S^c] = \frac{\Pr[S^c \mid V^c]\Pr[V^c]}{\Pr[S^c \mid V^c]\Pr[V^c] + \Pr[S^c \mid V]\Pr[V]}.$$ I leave the computation to you.
Another way to do these calculations is to construct a frequency table based on the given probabilities for a hypothetical cohort. Suppose the population contains $10000$ people. Of these, $(0.04)(10000) = 400$ are infected with the virus. The remaining $9600$ are healthy. Of the $400$ infected people, the test is $95\%$ reliable, so $(0.95)(400) = 380$ of these will test positive and $20$ will test negative. Of the $9600$ healthy people, the test is $99\%$ reliable, so $(0.99)(9600) = 9504$ of these will test negative, and $9600-9504 = 96$ will test positive. In summary
$$\begin{array}{|c|c|c|c|}
\hline
& V & V^c & \text{Total} \\
\hline
S & 380 & 96 & 476 \\
\hline
S^c & 20 & 9504 & 9524 \\
\hline
\text{Total} & 400 & 9600 & 10000 \\
\hline
\end{array}$$
We simply populated the corresponding joint events with the number of people we expect to meet the criteria, and the row totals $476$ and $9524$ were just the sums of the corresponding rows.
Now that we have constructed such a table, it is immediately obvious that $$\Pr[V \mid S] = \frac{380}{476} \approx 0.798319,$$ and the computation of $\Pr[V^c \mid S^c]$ is similarly performed by reading the appropriate cells in the table.
