Biased coin with a $3/4$ chance to land on the side it was before the flip

Let $p_n$ be the probability the coin is on tails after $n$ flips. Note that $p_0=0$.

The coin can be on tails after $n+1$ flips in two different ways: (i) it was on tails after $n$ flips, and the next result was a tail or (ii) it was on heads after $n$ flips, and the next result was a tail.

The probability of (i) is $(3/4)p_n$ and the probability of (ii) is $(1/4)(1-p_n)$. Thus $$p_{n+1}=\frac{1}{4}+\frac{1}{2}p_n.\tag{1}$$ Solve this recurrence relation. The general solution of the homogeneous recurrence $p_{n+1}=\frac{1}{2}p_n$ is $A\cdot \frac{1}{2^n}$. A particular solution of the recurrence (1) is $\frac{1}{2}$. So the general solution of the recurrence (1) is $$p_{n}=A\cdot \frac{1}{2^n}+\frac{1}{2}.$$ Set $p_0=0$ to find $A$. We find that $p_n=\frac{1}{2}-\frac{1}{2^{n+1}}$.

If you mean that $\Pr(H_i|H_{i-1}) = \Pr(T_i|T_{i-1}) = 3/4$ for all $i$, then this is a two-state homogeneous Markov chain. The transition matrix is

$$ P = \begin{pmatrix} 3/4 & 1/4 \\ 1/4 & 3/4 \end{pmatrix}. $$

This should get you started....

I'd comment but I don't have the rep, strangely you require 50.

Also, with regards to the bias in a coin flip, it's a paper by Diaconis:

http://statweb.stanford.edu/~susan/papers/headswithJ.pdf

Let $p_n$ be the probability that the coin is heads up after $n$ tosses. Then $p_0=1$ since it starts on heads, and $$p_n=\frac34p_{n-1}+\frac14(1-p_{n-1})$$ which simplifies to $$\left(p_n-\frac12\right)=\frac12\left(p_{n-1}-\frac12\right)\ .$$ Iterating, $$p_n-\frac12=\Bigl(\frac12\Bigr)^{n+1}$$ so $$p_n=\frac12+\Bigl(\frac12\Bigr)^{n+1}\ .$$