Calculate probability $P(\min\left\{X,Y\right\} \leq x)$ and $P(\max\left\{X,Y\right\} \leq x)$

$$\begin{split}F_{\min}(x) &= P(\min\left\{X,Y\right\} \leq x) \\[0.5ex] &= 1-P(x<\min\left\{X,Y\right\} ) \\[0.5ex] &= 1-P(x<X, x<Y) \\[0.5ex] & = 1-P(X>x)\,P(Y>x)\\[0.5ex] & = 1-(1-P(X \leq x))\,(1-P(Y \leq x))\\[0.5ex] & = 1-(1-F_X(x))\,(1-F_Y(x))\end{split}$$

Is this correct for minimum? I'm not sure how do it for $\max$?

It is correct, and for max it is even simpler.

$$\begin{split}F_{\max}(x) &= P(\max\left\{X,Y\right\} \leq x) \\[0.5ex] &= P(X\leq x, Y\leq x) \\[0.5ex] & = P(X\leq x)\,P(Y\leq x)\\[0.5ex] & = F_X(x)\,F_Y(x)\end{split}$$

Also, recall the expansion of a Geometric Series.

$$\displaystyle F_X(x)~{=F_Y(x)\\[0.5ex]= (\sum_{k=1}^x {2}^{-k})\;\mathbf 1_{x\in\{1,2,\ldots\}} \\[0.75ex]= (1-{2}^{-x})\;\mathbf 1_{x\in\{1,2,\ldots\}}}$$