(a) \[\left(\frac{4}{25}\right)^{-\frac12}=\left(\frac{25}{4}\right)^{\frac12}=\frac52,\qquad 2^{4}=16,\qquad \left(\frac{15}{2}\right)^{-2}=\left(\frac{2}{15}\right)^{2}=\frac{4}{225}.\] So \[\frac52\times16\div\frac{4}{225}=40\times\frac{225}{4}=2250.\]
(b) \[\log_{5}\!\frac35+3\log_{5}\!\frac52-\log_{5}\!\frac{81}{8}=\log_{5}\!\left(\frac35\cdot\left(\frac52\right)^{3}\cdot\frac{8}{81}\right)=\log_{5}\!\left(\frac35\cdot\frac{125}{8}\cdot\frac{8}{81}\right)=\log_{5}\!\frac{25}{27}.\] Hence the value is \[\log_{5}\frac{25}{27}=2-3\log_{5}3\approx-0.048.\]
(a) \[\left(\frac{4}{25}\right)^{-\frac12}=\left(\frac{25}{4}\right)^{\frac12}=\frac52,\qquad 2^{4}=16,\qquad \left(\frac{15}{2}\right)^{-2}=\left(\frac{2}{15}\right)^{2}=\frac{4}{225}.\] So \[\frac52\times16\div\frac{4}{225}=40\times\frac{225}{4}=2250.\]
(b) \[\log_{5}\!\frac35+3\log_{5}\!\frac52-\log_{5}\!\frac{81}{8}=\log_{5}\!\left(\frac35\cdot\left(\frac52\right)^{3}\cdot\frac{8}{81}\right)=\log_{5}\!\left(\frac35\cdot\frac{125}{8}\cdot\frac{8}{81}\right)=\log_{5}\!\frac{25}{27}.\] Hence the value is \[\log_{5}\frac{25}{27}=2-3\log_{5}3\approx-0.048.\]