Capacity of Binary Erasure Channel
By symmetry, we can assume that $\Pr[X=0] = \Pr[X=1] = p$, and so $\Pr[X=?] = 1-2p$. We can then express $I(X;Y)$ as a function of $p,\lambda,\mu$. The capacity of the channel is the maximum of $I(X;Y)$ over all $p \in [0,1/2]$. You can find this maximum using calculus.
In more detail, let us first note that $\Pr[Y = 0] = \Pr[Y = 1] = (1-\mu)p$ and $\Pr[Y=?] = 1-2(1-\mu)p$. Furthermore, $H(Y|X=?) = 0$ and $H(Y|X=0) = H(Y|X=1) = H(1-\lambda-\mu,\lambda,\mu)$. Therefore
$$
I(X;Y) = H((1-\mu)p,(1-\mu)p,1-2(1-\mu)p) - 2pH(1-\lambda-\mu,\lambda,\mu).
$$
The derivative of this with respect to $p$ is
$$
-2(1-\mu)\log((1-\mu)p)+2(1-\mu)\log(1-2(1-\mu)p) - 2H(1-\lambda-\mu,\lambda,\mu) = \\
2(1-\mu) \log \frac{1-2(1-\mu)p}{(1-\mu)p} - 2H(1-\lambda-\mu,\lambda,\mu).
$$
Equating this to zero, we get
$$
\log \frac{1-2(1-\mu)p}{(1-\mu)p} = \frac{H(1-\lambda-\mu,\lambda,\mu)}{1-\mu}
$$
and so
$$
\frac{1-2(1-\mu)p}{(1-\mu)p} = e^{H(1-\lambda-\mu,\lambda,\mu)/(1-\mu)}.
$$
Massaging this, we finally get
$$
p = \frac{1}{(1-\mu)(2 + e^{H(1-\lambda-\mu,\lambda,\mu)/(1-\mu)})}.
$$