解:
(1)$\because$双曲线$y =-\frac{4}{x}$过点$(a,2),$
$\therefore 2a=-4,$解得$a = - 2.$
又$\because$直线$y = kx$过点$(-2,2),$
$\therefore - 2k = 2,$解得$k = - 1.$
(2)$\because$将直线$y=-x$向上平移$m(m > 0)$个单位长度后对应的函数表达式为$y=-x + m,$$\therefore P(0,m),$$E(m,0),$
从而$OP = OE = m.$
过点$C$作$CF\perp y$轴于点$F,$则$\angle CFP=\angle EOP = 90^{\circ},$

$\because\angle CPF=\angle EPO,$$PE = PC,$
$\therefore\triangle PFC\cong\triangle POE(AAS).$
$\therefore PO = PF = m,$$CF = OE = m.$$\therefore C(-m,2m).$
$\because$双曲线$y =-\frac{4}{x}$过点$C(-m,2m),$
$\therefore - 2m^{2}=-4,$解得$m=\pm\sqrt{2}$(负值舍去).
$\therefore m$的值为$\sqrt{2}.$