I'm getting stuck with the diffusion equation
$k∂^2u/∂x^2=∂u/∂t$
with $0 < x < 1$, $t > 0$ and
$u(0, t) = 0,$
$∂u/∂x(1,t)= −hu(1, t),$
$h > 0,$ $t > 0$
$u(x, 0) = 1,$
$0 < x < 1.$
So far I've got my answer into the form $u(x,t)=\sum_{n=1}^{+\infty}C_n e^{-\lambda_n^2t}\sin(\mu_nx)$ (not 100% sure this is correct). But now I am struggling to find a value for $C_n$. I think I need to use Fourier inversion but I'm not overly familiar with using this.
Your series expansion is basically correct, but it should be said that $\mu_n$ come from the equation $-h\tan \mu = \mu$, which expresses the boundary condition at $x=1$. The infinite sequence of solutions of this equation, $\mu_n$, does not admit a close form.
Next, $\lambda_n$ in the exponent should be related to $\mu_n$, by way of the PDE. Plugging each term of the series into the PDE shows $\lambda_n^2 = k\mu_n^2$, so there isn't a need to write $\lambda_n$ at all: $$ u(x, t) = \sum_{n=1}^\infty C_n e^{-k\mu_n^2t}\sin\mu_n x \tag1 $$ To find $C_n$, we have to expand the initial condition $u(x,0)\equiv 1$ in the orthogonal basis $\sin \mu_n x$. (The orthogonality follows from the general theory of eigenfunctions with certain boundary conditions.) This yields $$ C_n = \frac{\int_0^1 1\sin \mu_n x\,dx}{\int_0^1 \sin^2 \mu_n x\,dx} = \frac{\mu_n^{-1}(1-\cos \mu_n)}{\frac12 - \frac1{4\mu_n}\sin2\mu_n} = 4 \frac{1 - \cos \mu_n}{2\mu_n - \sin 2\mu_n} $$ We don't get an explicit formula for $C_n$ since there isn't one for $\mu_n$. Asymptotically, $\mu_n$ is like $\pi n$ and so $C_n = O(1/n)$ as $n\to\infty$.