The$\newcommand{\Cov}{\operatorname{Cov}}$The multiplication rule for covariances is $Cov(aX,bY)=abCov(X,Y)$$\Cov(aX,bY)=ab\Cov(X,Y)$. Note that the constant term ($-\frac{160}{9}$) doesn't affect the standard deviation or covariances of the temperatures, so we can ignore it. Since $T=\frac{5}{9}X$ and $S=\frac{5}{9}Y$, we know that $$Cov(T,S)=Cov(\frac{5}{9}X,\frac{5}{9}Y)=(\frac{5}{9})^2 Cov(X,Y)=(\frac{5}{9})^2 4$$$$\Cov(T,S)=\Cov\left(\frac{5}{9}X,\frac{5}{9}Y\right)=(\frac{5}{9})^2 \Cov(X,Y)=\left(\frac{5}{9}\right)^2 4$$
The multiplication rule for standard deviations comes from the rule for variances: $s_{kX}=ks_X$. So, $$\rho(T,S)=\frac{Cov(T,S)}{s_{T} s_S}=\frac{Cov(\frac{5}{9}X,\frac{5}{9}Y)}{s_{\frac{5}{9}X}s_{\frac{5}{9}Y}}=\frac{(\frac{5}{9})^2 Cov(X,Y)}{(\frac{5}{9})^2 s_X s_Y}=\frac{Cov(X,Y)}{s_X s_Y} = \rho(X,Y)=0.8$$$$\rho(T,S)=\frac{\Cov(T,S)}{s_T s_S}=\frac{\Cov(\frac{5}{9}X,\frac{5}{9}Y)}{s_{\frac{5}{9}X}s_{\frac{5}{9}Y}}=\frac{(\frac{5}{9})^2 \Cov(X,Y)}{(\frac{5}{9})^2 s_X s_Y}=\frac{\Cov(X,Y)}{s_X s_Y} = \rho(X,Y)=0.8$$
