The problem assumes that the probability of the $i$-th adoption being a dog, or a cat is $\frac{1}{2}$. The problem also assumes that the events that the $i$-th and the $j$-th adoption are statistically independent for $i\neq j$. If the pet adopter plans to have $n$ pets and $A$ denotes the event that they have both a dog and a cat, then

\[A^C=\{\underbrace{( DDD\ldots D)}_{n}, \underbrace{ (CCC\ldots C)}_{n}\}.\]

Let us note that this event the complement of A is made up of two atomic events all dogs and all cats. The event is a union of atomic events, an or relationship. Secondly, by construction they are independent.

So, using the addition rule and the independence assumption, we get\[P(A^C)=P(\underbrace{DDD\ldots}_{n})+P( \underbrace{CCC\ldots}_{n})=\left(\frac{1}{2}\right)^n+\left(\frac{1}{2}\right)^n=2\left(\frac{1}{2}\right)^n.\]

Now, we have\[\begin{align}&P(A)\geq0.95\\
\implies&1-P(A^C)\geq0.95\\
\implies&P(A^C)\leq0.05\\
\implies&2\left(\frac{1}{2}\right)^n\leq0.05\\
\implies&\left(\frac{1}{2}\right)^n\leq0.025\\
\implies&n\log{\left(\frac{1}{2}\right)}\leq\log{0.025}\\
\implies&n\geq5.32.\end{align}\]

Therefore, the pet-adopter should plan to adopt at least $\lceil n\rceil=6$ pets.