Answer
a) $$\frac{e^8-1}{2e^4}$$
b) $$\frac{15}{8}$$
Work Step by Step
Recall that:
$$\sinh x =\frac{e^x-e^{-x}}{2}$$
a) $\sinh 4$
$=\frac{e^4-e^{-4}}{2}
=\frac{e^4-\frac{1}{e^4}}{2}
=\frac{\frac{e^8-1}{e^4}}{2}
=\frac{e^8-1}{2e^4}$
b) $\sinh ({\ln4})$
$=\frac{e^{\ln 4}-e^{-({\ln 4})}}{2}
=\frac{4-\frac{1}{4}}{2}
=\frac{\frac{15}{4}}{2}
=\frac{15}{8}$