1. The line x + y = k has a slope of -1. Therefore, the line perpendicular to it that passes through the center of the circle x^2 + y^2 = k has a slope of 1. The center of the circle is (0, 0), so the equation of the line perpendicular to x + y = k that passes through (0, 0) is y = x. Solving the system of equations x + y = k and x^2 + y^2 = k, we get two solutions: (k/2, k/2) and (-k/2, -k/2). Since the line y = x is tangent to the circle at (k/2, k/2), we have k/2 + k/2 = k, which gives k = 2.
2. The line y = 1/2x + 5 has slope 1/2, so the line perpendicular to it that passes through A = (6, 2) has slope -2. The equation of this line is y - 2 = -2(x - 6), which simplifies to y = -2x + 14. To find the reflection B of A over the line y = 1/2x + 5, we need to find the intersection of the line y = -2x + 14 and the perpendicular bisector of the segment AB. The midpoint of AB is ((6 + x)/2, (2 + y)/2), and the slope of AB is (y - 2)/(x - 6). Therefore, the slope of the perpendicular bisector of AB is -1/(y - 2)/(x - 6), which simplifies to (6 - x)/(y - 2). The equation of the perpendicular bisector of AB is y - (2 + y)/2 = (6 - x)/(y - 2)(x - 6)(x - (6 + x)/2) + (2 + y)/2. Solving the system of equations y = -2x + 14 and y - (2 + y)/2 = (6 - x)/(y - 2)(x - 6)(x - (6 + x)/2) + (2 + y)/2, we get B = (5, 8). Therefore, the coordinates of B are (5, 8).