Answer
See below for proofs.
Work Step by Step
$(a)$
We prove:
$\sinh(x+y) =\sinh x\cosh y+\cosh x\sinh y\qquad (*)$
$\displaystyle \sinh x\cosh y=\frac{e^{x}-e^{-x}}{2}\cdot\frac{e^{y}+e^{-y}}{2}=\frac{e^{x+y}+e^{x-y}-e^{y-x}-e^{-y-x}}{4}$
$\displaystyle \cosh x\sinh y=\frac{e^{x}+e^{-x}}{2}\cdot\frac{e^{y}-e^{-y}}{2}=\frac{e^{x+y}-e^{x-y}+e^{y-x}-e^{-x-y}}{4}$
Adding, we have the RHS of equation (1)
RHS=$\displaystyle \frac{2e^{x+y}-2e^{-(x+y)}}{4}=\frac{e^{(x+y)}-e^{-(x+y)}}{2}=\sinh(x+y)$
The first formula is valid.
To show that (a) is valid,
$\sinh(2x)=\sinh(x+x)=\qquad $apply (*)
$=\sinh x\cosh x+\cosh x\sinh x$
$=2\sinh x\cosh x$
$(b)$
We prove:
$\cosh(x+y) =\cosh x\cosh y+\sinh x\sinh y \quad(**)$
$RHS=\displaystyle \frac{e^{x}+e^{-x}}{2}\cdot \frac{e^{y}+e^{-y}}{2}+\frac{e^{x}-e^{-x}}{2}\cdot\frac{e^{y}-e^{-y}}{2}$
$=\displaystyle \frac{e^{x+y}+e^{x-y}+e^{-x+y}+e^{-x-y}}{4}+\frac{e^{x+y}-e^{x-y}-e^{-x+y}+e^{-x-y}}{4}$
$=\displaystyle \frac{2e^{x+y}+2e^{-(x+y)}}{4}$
$=\displaystyle \frac{e^{x+y}+e^{-(x+y)}}{2}=\cosh(x+y)$, so the second formula is valid.
To show that (b) is valid,
$\cosh(2x)=\cosh(x+x)=\qquad $apply ($**$)
$=\cosh x\cosh x+\sinh y\sinh y$
$=\cosh^{2}x+\cosh^{2}y$
