Probability of second card being an ace

As Ant said in the comments, in both cases, the answer is $\frac{4}{52} = \frac{1}{13}$. Your first equation simplifies to this: \begin{align*} \left(\frac{4\cdot 3}{52 \cdot 51}\right) + \left(\frac{48 \cdot 4}{52 \cdot 51}\right) &= \left(\frac{4}{52}\right)\left( \frac{3}{51} + \frac{48}{51}\right) \\ &= \left(\frac{4}{52}\right)(1) \\ &= \frac{1}{13} \end{align*} But this is actually overcomplicating things; drawing the first card doesn't affect the odds at all, regardless of whether or not it's replaced. We don't know anything about what the first card drawn was, so it doesn't add any information, and thus doesn't affect the probability that the second card is an ace.

From a probabilistic point of view, extracting a card and placing it on same table of your deck without looking at it is like not extracting it at all: you have the same information you had before. More precisely given a probability space $(\Omega,\mathcal F,\mathbb P)$ your $\sigma$-algebra $\mathcal F$ is not changed, so your probability space is not changed.

Mathematically if $E_1=\{$Ace at the first extraction$\}$ and $E_2=\{$Ace at the second extraction$\}$ then $$P(E_2)=P(E_2\cap E_1)+P(E_2\cap E_1^c)$$ because $E_1$ and $E_1^c$ are a partition of $\mathcal F$.

From the definition of conditional probability $P(E_2|E_1)=\frac{P(E_2\cap E_1)}{P(E_1)}$, so $P(E_2\cap E_1)=P(E_2|E_1)P(E_1)$ and $P(E_2\cap E_1^c)=P(E_2|E_1^c)P(E_1^c)$.

Now, know the first card was an ace, means that you have 51 cards in your deck now and only 3 aces in that, so $P(E_2|E_1)=\frac 3 {51}$ and obviously $P(E_1)=\frac 4 {52}$. Similarly, knowing your first card wasn't an ace means you have 4 aces in a deck of 51 cards , so $P(E_2|E_1^c)=\frac 4 {51}$ and $P(E_1^c)=1-P(E_1)=\frac {48} {52}$. So $$P(E_2)=P(E_2\cap E_1)+P(E_2\cap E_1^c)\\=\frac 3 {51}\cdot\frac 4 {52}+\frac 4 {51}\cdot\frac {48} {52}\\=\frac 4 {52}[\frac 3 {51}+\frac {48} {51}]\\=\frac 4 {52}=\frac 1 {13} $$