Define $X$ as the following random variable $$\begin{array}{r}X=\{\begin{array}{ll}1,p=\frac{1}{4}& HHH\text{ or }TTT\\ 0,p=\frac{3}{4}& \text{otherwise}\end{array}\end{array}$$

1. $K$ is simply a binomial distribution, where we want the $2^{nd}$ success to happen at the $K+1$th trial. $$\begin{array}{r}{p}_K(k)=(\genfrac{}{}{0ex}{}{k}{1}){\frac{1}{4}}^2{\frac{3}{4}}^{k-1}\text{since the last trial is success}\end{array}$$

2. $M$ = number of tails before first success. Let the success be at $N+1$. Defin $Y$ as $$\begin{array}{rl}Y& =\{\begin{array}{ll}1p=\frac{1}{2}& HHT\text{, }HTH\text{, or }THH\\ 2p=\frac{1}{2}& HTT\text{, }THT\text{, or }TTH\end{array}\\ E[Y]& =1\times \frac{1}{2}+2\times \frac{1}{2}\\ Var(Y)& =(1-\frac{3}{2}{)}^2\times \frac{1}{2}+(2-\frac{3}{2}{)}^2\times \frac{1}{2}\\ E[N+1]& =\frac{1}{p}=4\\ Var(N+1)& =Var(N)=\frac{1-p}{p^2}=\frac{1-\frac{1}{4}}{{\frac{1}{4}}^2}\\ M& =Y_1+Y_2+\cdots Y_N\\ E[M]& =E[Y_1+Y_2+\cdots Y_N]\\ Var(M)& =Var(Y_1+Y_2+\cdots Y_N)\end{array}$$ Note that both $Y$ and $N$ are random variables here. Using the formulae for random number of random variables, $$\begin{array}{rl}E[M]& =E[E[M|N]]=E[NE[Y]]=E[N]E[Y]=(4-1)\times \frac{3}{2}=\frac{9}{2}\\ Var(M)& =Var(E[M|N])+E[Var(M|N)]=Var(NE[Y])+E[NVar(Y)]\\ & =E[Y{]}^{2}Var(N)+E[N]Var(Y)=\frac{9}{4}\times 12+3\times \frac{1}{4}=\frac{111}{4}\end{array}$$