I'm surprised how many bad, suboptimal and/or overcomplicated answers this question has inspired, when there's a fairly simple solution; and that the only answer that mentions and uses the most relevant fact, Christian Blatter's, didn't have a single upvote before I just upvoted it.
The $2$-sphere is unique in that slices of equal height have equal surface area. That is, to sample points on the unit sphere uniformly, you can sample $z$ uniformly on $[-1,1]$ and $\phi$ uniformly on $[0,2\pi)$. If your cone were centred around the north pole, the angle $\theta$ would define a minimal $z$ coordinate $\cos\theta$, and you could sample $z$ uniformly on $[\cos\theta,1]$ and $\phi$ uniformly on $[0,2\pi)$ to obtain the vector $(\sqrt{1-z^2}\cos\phi,\sqrt{1-z^2}\sin\phi,z)$ uniformly distributed as required.
So all you have to do is generate such a vector and then rotate the north pole to the centre of your cone. If the cone is already thus centred, you're done; if it's centred on the south pole, just invert the vector; otherwise, take the cross product of the cone's axis with $(0,0,1)$ to get the direction of the rotation axis, and the scalar product to get the cosine of the rotation angle. Or if you prefer you can apply your idea of generating two orthogonal vectors, in the manner Christian described.
First will come the pretty complete theory. Then we look at your particular situation.
Theory: We need an expression for the variance of $X$. The variance is $E(X^2)-(E(X))^2$. For $E(X^2)$, we need to calculate
$$\int_a^b \frac{x^2}{b-a}\,dx,$$
which is $\frac{b^3-a^3}{3(b-a)}$. This simplifies to $\frac{b^2+ab+a^2}{3}$.
I imagine that you know that $E(X)=\frac{b+a}{2}$. One can do this by integration, but it is clear by symmetry that the mean is halfway between $a$ and $b$.
So we know that the variance is $\frac{b^2+ab+a^2}{3}-\frac{(b+a)^2}{4}$. Bring to a common denominator, simplify. We get that
$$\text{Var}(X)=\frac{(b-a)^2}{12} \tag{$\ast$}.$$
More simply, you can search under uniform distribution, say on Wikipedia. They will have the expression $(\ast)$ for the variance of $X$.
Your problem: We know that $\frac{b+a}{2}=3.5$. We also know that the standard deviation of $X$ is $1.3$, so the variance is $(1.3)^2=1.69$.
So, by $(\ast)$, $\frac{(b-a)^2}{12}=1.69$, and therefore $b-a=\sqrt{(12)(1.69)}\approx 4.5033$. We also know that $b+a=(2)(3.5)=7$. Now that we know $b-a$ and $b+a$, it is easy to find $a$ and $b$.
For your simulation, presumably you are starting from a random number generator that generates numbers that are more or less uniformly distributed on $[0,1)$. If $U$ represents the output of such a generator, we simulate $X$ by using $a+(b-a)U$. And we now know $a$ and $b$.
Added If you want a general formula instead of a procedure, let $\mu=\frac{a+b}{2}$ be the mean, and $\sigma=\frac{b-a}{\sqrt{12}}=\frac{b-a}{2\sqrt{3}}$ be the standard deviation. Then $\frac{b-a}{2}=\sqrt{3}\,\sigma$.
We get $a=\frac{b+a}{2}-\frac{b-a}{2}=\mu-\sqrt{3}\,\sigma$. It follows that we can take
$$X=\mu-\sqrt{3}\,\sigma + (2\sqrt{3}\,\sigma) U=\mu+(\sqrt{3}\,\sigma)(2U-1).$$
Best Answer
Note that this is the strategy for random points in a unit ball, not for random points on a unit sphere.
Let us assume you have generated $v$ as a Gaussian $d$-dimensional vector. Normalizing this to length $1$ is done by using $\frac{v}{\|v\|}$ as you correctly stated. This results in a uniform distribution on the surface of the unit sphere, as described in the article about hypersphere point picking.
In order to get points inside of the ball, you have to use another random variable to specify the distance from the origin. If you simply used a uniform distribution for this, you would get too many points near the origin and not enough points near the surface.
But if you take a variable $u$ with a uniform distribution and raise it to the $\frac{1}{d}$-th power, the resulting distribution ensures that the distribution of points inside the ball is uniform. In a manner of speaking, applying the $\frac{1}{d}$-th power moves the points outwards by just the correct amount to get a uniform distrubution in the ball.