Part 1:

Write e^x=x^2+k as e^x-x^2=k.

The derivative of e^x - x^2 is e^x - 2x. If there were a maximum or minimum, this continuous function should have a zero in the derivative, or e^x=2x.  Does this derivative have a zero? For negative x, the LHS is positive and the RHS is negative, so e^x exceeds 2x and at x=0 this is still true. The original function is still increasing.

The second derivative, e^x-2 becomes zero (from negative to positive) at ln(2). At this point the first derivative is 2 - 2 ln(2), which, since ln(2) is less than 1, is still positive. Since the curve now begins to be concave up, the slope will remain positive from then on and the function will continue to rise and there will be only one real root regardless of the level at which k is placed.

Part 2:

Graphing shows a local maximum of y = e^x - x^4 near x = 0.82, with y reaching about 1.81, so that's approximately the highest value of k giving more than one root.

Graphing also shows a local minimum near x = 7.3 where y is about -1370, so that's approximately the lowest value of k giving multiple real roots.

The derivative of e^x - x^4 is e^x - 4x^3, so we need to find the two x values that zero this expression.

This is probably too hard to evaluate algebraically, and rather than write a program to home in on an approximation, I asked Wolfram Alpha

x ~= 0.8310314521082453293

x ~= 7.3844056621960990434

For the former, e^x - x^4 ~= 1.818738714288628143, and for the latter, e^x - x^4 ~= -1362.790360110545366.

Between these two values of k there are 3 real solutions.