First, it's not hard to show that
$$u(x) = a\sin(\sqrt{\lambda}\,x) + b\cos(\sqrt{\lambda}\,x)$$
is the general solution of the differential equation. Thus,
$$u'(x) = a\sqrt{\lambda}\cos(\sqrt{\lambda}\,x) - b\sqrt{\lambda}\sin(\sqrt{\lambda}\,x)$$
so left hand boundary condition becomes
$$u'(0) = a\sqrt{\lambda}\cos(\sqrt{\lambda}\,0) - b\sqrt{\lambda}\sin(\sqrt{\lambda}\,0) = a\sqrt{\lambda} = 0.$$
Thus, either $\lambda = 0$ or $a=0$. If $\lambda = 0$, then the solution is just $u(x)=b$ so that the right hand boundary condition is not satisfied. We are stuck with $a=0$. Thus, $u(x)$ simplifies to
$$u(x) = b\cos(\sqrt{\lambda}\,x)$$
and $u'(x)$ simplifies to
$$u'(x) = - b\sqrt{\lambda}\sin(\sqrt{\lambda}\,x).$$
Examining the right hand boundary condition, we get
$$u(1)+u'(1) = b\cos(\sqrt{\lambda}) - b\sqrt{\lambda}\sin(\sqrt{\lambda}) = 0.$$
Dividing off the $b$, this can be rewritten as
$$\tan(\sqrt{\lambda}) = 1/\sqrt{\lambda}$$
or, setting $p=\lambda$,
$$\tan(p)=1/p.$$
To be clear - this is a condition that $\lambda$ must satisfy in order for the problem to have a non-trivial solution. For any other choice of $\lambda$, $u(x)=0$ is the only solution but, when $\tan(\sqrt{\lambda}) = 1/\sqrt{\lambda}$, $u(x) = b\cos(\sqrt{\lambda}\,x)$ is a solution for every $b$.
The problem asks us to plot the functions $\tan(p)$ and $1/p$ together to illustrate the solutions. Here's how to do that with Mathematica (not that it's necessary to use a program).
Plot[{Tan[p], 1/p}, {p, 0, 10}, PlotPoints -> 100,
Exclusions -> Cos[p] == 0, ExclusionsStyle -> Dashed]
Note that the points of intersection can only be computed numerically - there is no closed form. We can find them like so:
solutions = p /. NSolve[Tan[p] == 1/p && 0 < p < 10, p]
(* Out: {0.860334, 3.42562, 6.4373, 9.52933} *)
The eigenvalues are the squares of these numbers
solutions^2
(* Out: {0.740174, 11.7349, 41.4388, 90.8082} *)
Here's what the graph of one of the solutions looks like.
u[x_] = Cos[3.425618 x];
Plot[u[x], {x, 0, 1}, AspectRatio -> Automatic]
It's pretty easy to see that the left hand boundary condition is satisfied. Note also that the value $u(1)$ looks very close to $-1$ while the slope $u'(1)$ looks very close to $1$. Thus, the right hand boundary condition looks satisfied as well. We can check!
{u[1], u'[1], u[1] + u'[1]}
(* Out: {-0.959935, 0.959933, -1.76846*10^-6} *)
