Answer :

[tex]k:y=m_1x+b_1;\ l:y=m_2x+b_@\\\\k\ \perp\ l\iff m_1m_2=-1\\\\========================\\k:y=x-1;\ l:y=mx+b\\\\k\ \perp\ l\iff1m=-1\to m=-1\\\\l:y=-1x+b\to y=-x+b\ (b\in\mathbb{R})\\==========================\\k:y=-\frac{1}{3}x-1;\ l:y=mx+b\\\\k\ \perp\ l\iff-\frac{1}{3}m=-1\to m=3\\\\l:y=3x+b\ (b\in\mathbb{R})[/tex]
AL2006
Those are two separate lines, and they're not parallel or perpendicular
to each other.
So there's no line that's perpendicular to both of them, and you're asking
two separate questions.

For both of them, you have to remember this: 
Lines that are perpendicular have negative reciprocal slopes.

-- The slope of [ y = x - 1 ] is 1, and the negative reciprocal of 1 is -1/1 = -1 .
A line perpendicular to [ y = x - 1 ] is [ y = -x + any number ].

-- The slope of [ y = -1/3 x - 1 ] is -1/3, and the negative reciprocal of -1/3 is 3 .
A line perpendicular to [ y = -1/3 x - 1 ] is [ y = 3x + any number ].


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